125876 + Interview Questions in General QA Page 629 InterviewSolution

31401.	आवृत्ति वितरण में विषमता है ऐसा कब कहा जाता है ?
Answer» संमितता या सुडोलता की कमीवाले आवृत्ति वितरण में विषमता है ऐसा कहेंगे ।

Discussion

31402.	एक आवृत्ति वितरण के तीन चतुर्थक 42, 36 और 40 है, तो आवृत्ति वितरण का प्रकार बताइए ।
Answer» यहा Q₁ = 36, Q₃ = 42, M = 40 होगा । Q₃ – M > M – Q₁ है इसलिए ऋण विषमतावाला आवृत्ति वितरण है ।

Discussion

31403.	संमित आवृत्ति वितरण किसे कहते है ?
Answer» यदि समष्टि के अवलोकन भूयिष्ठक के मूल्य से दोनों ओर समान रीति से वितरित हुए हो एसे आवृत्ति वितरण को संमित आवृत्ति वितरण कहते है।

Discussion

31404.	जब खुल्ला शिरावाला आवृत्ति वितरण और एक से अधिक भूयिष्ठक हो तब विषमतांक कौन से सूत्र से प्राप्त करोगे?
Answer» एक से अधिक भूयिष्ठकवाला आवृत्ति वितरण हो तब भूयिष्ठक के स्थान पर माध्यिका ज्ञात करके निम्न सूत्र का उपयोग करेंगे । \(j = \frac{3(\bar X−M)}S\)

Discussion

31405.	एक धन विषम आवृत्ति वितरण का विषमतांक 0.75 है । यदि प्र. विचलन 20 और माध्य 37.50 हो, तो M ज्ञात करो।
Answer» यहाँ j = 0.75, S = 20, \(\overline X\) = 37.50 है । ∴ j = \(\frac{3(\overline X−M)}S\) 0.75 = \(\frac{3(37.50−M)}{20} \frac{0.75×20}3\) = 37.50 – M 5 = 37.50 – M ∴ M = 37.50 – 5 ∴ M = 32.50

31405.

एक धन विषम आवृत्ति वितरण का विषमतांक 0.75 है । यदि प्र. विचलन 20 और माध्य 37.50 हो, तो M ज्ञात करो।

Answer»

यहाँ j = 0.75, S = 20, \(\overline X\) = 37.50 है ।

∴ j = \(\frac{3(\overline X−M)}S\)

0.75 = \(\frac{3(37.50−M)}{20} \frac{0.75×20}3\) = 37.50 – M

5 = 37.50 – M

∴ M = 37.50 – 5

∴ M = 32.50

Discussion

31406.	यदि किसी आवृत्ति वितरण में (Q3 – Q2) < (Q2 – Q1) हो, तो आवृत्ति वितरण का प्रकार बताइए ।
Answer» यदि किसी आवृत्ति वितरण में (Q₃ – Q₂) < (Q₂ – Q₁) हो, तो ऋण आवृत्ति वितरण होगा ।

Discussion

31407.	एक आवृत्ति वितरण में Q3 – Q2 = 2 (Q2 – Q1) हो, तो j ज्ञात करो ।
Answer» j = \(\frac{(Q_3−Q_2)−(Q_2−Q_1)}{(Q_3−Q_2)+(Q_2−Q_1)}\) Q₃ – Q₂ के स्थान पर 2 (Q₂ – Q₁) रखने पर ∴ j = \(\frac{2(Q_2−Q_1)−1(Q_2−Q_1)}{2(Q_2−Q_1)+1(Q_2−Q_1) }\)= \(\frac{1}2\) j = 0.33

31407.

एक आवृत्ति वितरण में Q3 – Q2 = 2 (Q2 – Q1) हो, तो j ज्ञात करो ।

Answer»

j = \(\frac{(Q_3−Q_2)−(Q_2−Q_1)}{(Q_3−Q_2)+(Q_2−Q_1)}\)

Q₃ – Q₂ के स्थान पर 2 (Q₂ – Q₁) रखने पर

∴ j = \(\frac{2(Q_2−Q_1)−1(Q_2−Q_1)}{2(Q_2−Q_1)+1(Q_2−Q_1) }\)= \(\frac{1}2\)

j = 0.33

Discussion

31408.	Which of the following statements is true for a symmetric distribution?(a) Q3 = 2M – Q1(b) Q2 – Q3 = Q2 – Q1(c) Q3 + Q1 > 2M(d) Q3 + Q1 < 2M
Answer» Correct option is (a) Q₃ = 2M – Q₁

Discussion

31409.	एक आवृत्ति वितरण की विषमता SK = – 2.8 है । यदि उसका भूयिष्ठक 48.8 हो, तो माध्य ज्ञात करो ।
Answer» SK =\( \overline X\) – M_O M_O = 48.8 दिया है । ∴ – 2.8 = \( \overline X\) – 48.8 ∴ – 2.8 + 48.8 = \( \overline X\) ∴\( \overline X\)= 46 माध्य = 46

31409.

एक आवृत्ति वितरण की विषमता SK = – 2.8 है । यदि उसका भूयिष्ठक 48.8 हो, तो माध्य ज्ञात करो ।

Answer»

SK =\( \overline X\) – M_O M_O = 48.8 दिया है ।

∴ – 2.8 = \( \overline X\) – 48.8

∴ – 2.8 + 48.8 = \( \overline X\)

∴\( \overline X\)= 46 माध्य = 46

Discussion

31410.	संमित आवृत्ति वितरण के लिए कौन-सा विधान सत्य है ? (A) Q3 = 2M – Q1(B) Q2 – Q3 = Q2 – Q1(C) Q3 + Q1 > 2M(D) Q3 + Q1 < 2M
Answer» सही विकल्प है (A) Q₃ = 2M – Q₁

Discussion

31411.	एक आवृत्ति वितरण के लिए Q3 + Q1 = 125 और M = 62.5 है, तो आवृत्ति वितरण की विषमता के बारे में क्या कह सकते हो ?
Answer» यहाँ Q₃ + Q₁ – 2M = 0 होता है । 125 – 2 × 62.5 = 125 – 125 = 0 इसलिए संमित आवृत्ति वितरण है ।

Discussion

31412.	एक आवृत्ति वितरण में Q3 + Q1 = 1.5 M और 3(Q3 – Q1) = 2M हो, तो j ज्ञात करो ।
Answer» यहाँ Q₃ + Q₁ = 1.5 M रखने पर और Q₃ – Q₁ =\(\frac{2M}3 \)रखने पर ∴ j = \(\frac{Q_3+Q_1−2M}{Q_3−Q_1}=\frac{1.5−2M}{0.67M}=\frac{−0.5M}{0.67M}\) ∴ j = – 0.75

31412.

एक आवृत्ति वितरण में Q3 + Q1 = 1.5 M और 3(Q3 – Q1) = 2M हो, तो j ज्ञात करो ।

Answer»

यहाँ Q₃ + Q₁ = 1.5 M रखने पर

और Q₃ – Q₁ =\(\frac{2M}3 \)रखने पर

∴ j = \(\frac{Q_3+Q_1−2M}{Q_3−Q_1}=\frac{1.5−2M}{0.67M}=\frac{−0.5M}{0.67M}\)

∴ j = – 0.75

Discussion

31413.	विषम आवृत्ति वितरण की परिभाषा दीजिए और उसके लक्षण बताइए ।
Answer» सममितियता का अभाव अर्थात् विषमता । उसके लक्षण निम्नलिखित है : जिस आवृत्ति वितरण का आवृत्तिवक्र संपूर्ण घंटाकार का न हो ऐसा आवृत्ति वितरण विषम है । जिस आवृत्ति वितरण का माध्य, मध्यका और भूयिष्ठक समान नहीं अर्थात् \(\overline X\) # M # M_O वह विषम आवृत्ति वितरण है । जिस आवृत्ति वितरण में चतुर्थक मध्यका से समान अन्तर पर न आये हो उस आवृत्ति वितरण विषम है ऐसा कहेंगे । . यदि मध्य में आये वर्ग का महत्तम आवृत्ति से दोनों ओर समान अन्तर पर आये वर्गों की आवृत्तियाँ समान न हो वह विषम आवृत्ति वितरण है । जिस आवृत्तिवक्र का आवृत्तिवक्र के दोनों किनार अधिक खिचा हुआ होता है ।

31413.

विषम आवृत्ति वितरण की परिभाषा दीजिए और उसके लक्षण बताइए ।

Answer»

सममितियता का अभाव अर्थात् विषमता । उसके लक्षण निम्नलिखित है :

जिस आवृत्ति वितरण का आवृत्तिवक्र संपूर्ण घंटाकार का न हो ऐसा आवृत्ति वितरण विषम है ।
जिस आवृत्ति वितरण का माध्य, मध्यका और भूयिष्ठक समान नहीं अर्थात् \(\overline X\) # M # M_O वह विषम आवृत्ति वितरण है ।
जिस आवृत्ति वितरण में चतुर्थक मध्यका से समान अन्तर पर न आये हो उस आवृत्ति वितरण विषम है ऐसा कहेंगे । .
यदि मध्य में आये वर्ग का महत्तम आवृत्ति से दोनों ओर समान अन्तर पर आये वर्गों की आवृत्तियाँ समान न हो वह
विषम आवृत्ति वितरण है ।
जिस आवृत्तिवक्र का आवृत्तिवक्र के दोनों किनार अधिक खिचा हुआ होता है ।

Discussion

31414.	Which of the following statements is false?(a) If Q3 + Q1 > 2M the distribution is positively skewed.(b) Bowley’s coefficient of skewness is found using positional averages.(c) The absolute measure is divided by standard deviation in Karl Pearson’s method to eliminate the effect of unit of the variable whereas in Bowley’s method, the absolute measure is divided by the difference of quartiles.(d) The frequencies of observations which are equidistant on both sides of the mode are equally distributed in a skewed distribution.
Answer» Correct option is (d) The frequencies of observations which are equidistant on both sides of the mode are equally distributed in a skewed distribution.

Discussion

31415.	आवृत्ति वितरण में यदि Q3 + Q1 = 60 और M = 30 हो, तो उसकी विषमता के लिए कौन-सा विधान सत्य है :(A) आवृत्ति वितरण अधिक विषम है।(B) आवृत्ति वितरण कम विषम है।(C) आवृत्ति वितरण में संमितता की कमी है।(D) आवृत्ति वितरण संमित है ।
Answer» सही विकल्प है (D) आवृत्ति वितरण संमित है ।

Discussion

31416.	Write the formula of quartile deviation.
Answer» Coefficient of quartile deviation = (Q₃ - Q₁)/(Q₃ + Q₁) Here, Q₃ = upper quartile, Q₁ = Lower quartile.

31416.

Write the formula of quartile deviation.

Answer»

Coefficient of quartile deviation

= (Q₃ - Q₁)/(Q₃ + Q₁)

Here, Q₃ = upper quartile, Q₁ = Lower quartile.

Discussion

31417.	Algebraic sum of deviations from mean is :(A) Negative(B) Positive(C) Different in each(D) Zero
Answer» Answer is (C) Different in each

Discussion

31418.	Coefficient of Range can be defined as –(A) (H - L)/2(B) (H + L)/2(C) (H - L)/(H + L)(D) (H + L)/(H - L)
Answer» Answer is (C) (H - L)/(H + L)

Discussion

31419.	Temperature of 7 days in a city are given in centigrade 18, 12, 6, -7, -12, 5, -4. Then range in centrigrade will be –(A) 6(B) 30(C) 22(D) 14
Answer» Answer is (B) 30 Range = Max. Value – Min. Value = 18 – (- 12) = 18 + 12 = 30

31419.

Temperature of 7 days in a city are given in centigrade 18, 12, 6, -7, -12, 5, -4. Then range in centrigrade will be –(A) 6(B) 30(C) 22(D) 14

Answer»

Answer is (B) 30

Range = Max. Value – Min. Value

= 18 – (- 12)

= 18 + 12 = 30

Discussion

31420.	Find the coefficient of range of the following data –Size10-1515-2020-2525-30Frequency2468
Answer» Lower limit (S) = 10 Upper limit (L) = 30 Range = (L - S)/(L + S) = (30 - 10)/(30 + 10) = 20/40 = 1/2 = 0.5 Thus Range coefficient is 0.5.

31420.

Find the coefficient of range of the following data –Size10-1515-2020-2525-30Frequency2468

Answer»

Lower limit (S) = 10

Upper limit (L) = 30

Range = (L - S)/(L + S)

= (30 - 10)/(30 + 10)

= 20/40

= 1/2 = 0.5

Thus Range coefficient is 0.5.

Discussion

31421.	Find the formula to find standard deviation in individual series.
Answer» σ = √((∑x²)/n - ((∑x)/n)²)

Discussion

31422.	If maximum cost of any times is Rs 500 and minimum Rs 75, then coefficient of range will be –(A) 0.739(B) 0.937(C) 7.39(D) 73.9
Answer» Answer is (A) 0.739 Range coefficient = (500 - 75)/(500 + 75) = 425/575 = 0.739

31422.

If maximum cost of any times is Rs 500 and minimum Rs 75, then coefficient of range will be –(A) 0.739(B) 0.937(C) 7.39(D) 73.9

Answer»

Answer is (A) 0.739

Range coefficient

= (500 - 75)/(500 + 75) = 425/575

= 0.739

Discussion

31423.	If value of all the terms of a series are same, then find value of dispersion.
Answer» Value of dispersion is zero because dispersion (deviation) cannot be find for same values.

Discussion

31424.	If variance (σ) = √(∑fd2/k - (∑fd/30)2) then value of k is –(A) 10(B) 20(C) 30(D) 60
Answer» Answer is (C) 30 Formula (σ) = √(∑fd²/k - (∑fd/30)²) k = 30

31424.

If variance (σ) = √(∑fd2/k - (∑fd/30)2) then value of k is –(A) 10(B) 20(C) 30(D) 60

Answer»

Answer is (C) 30

Formula (σ) = √(∑fd²/k - (∑fd/30)²)

k = 30

Discussion

31425.	Coefficient of range of variable series 10, 20, 30, 40, 50, 60 is –(A) 3/2(B) 5/6(C) 7/5(D) 5/7
Answer» Answer is (D) 5/7 Coefficient of Range = (60 - 10)/(60 + 10) = 50/70 = 5/7

31425.

Coefficient of range of variable series 10, 20, 30, 40, 50, 60 is –(A) 3/2(B) 5/6(C) 7/5(D) 5/7

Answer»

Answer is (D) 5/7

Coefficient of Range

= (60 - 10)/(60 + 10) = 50/70 = 5/7

Discussion

31426.	Marks obtain by five students in math are 20, 25, 15, 35 and 30 its range will be :(A) 15(B) 20(C) 25(D) 30
Answer» Answer is (B) 20 Range = Maximum value – minimum value = 35 – 15 = 20

31426.

Marks obtain by five students in math are 20, 25, 15, 35 and 30 its range will be :(A) 15(B) 20(C) 25(D) 30

Answer»

Answer is (B) 20

Range = Maximum value – minimum value

= 35 – 15 = 20

Discussion

31427.

Evaluate out range and coefficient of range of the marks.Marks5−910−1415−1920−2425−2930−34No. Students463264

Answer»

To get the range and its coefficient, first we have to convert inclusive class intervals into exclusive class intervals.

Class Interval	Exclusive Class Interval	Frequency
5 − 9	4.5 − 9.5	4
10 −14	9.5 − 14.5	6
20 −24	19.5 − 24.5	2
25 −29	24.5 − 29.5	6
30− 34	29.5 − 34.5	4

Range = Upper limit of the highest class interval − Lower limit of the lowest class intervalor, Range = 34.5 − 4.5 = 30

Coefficient or Range = \(\frac{H-L}{H+L}=\frac{34.5-4.5}{34.5+4.5}=\frac{30}{39}\) = 0.769

Hence, the range is 730 marks and coefficient of range is 0.769.

Discussion

31428.

From the following data calculate range and coefficient of range:Marks15253545556575No. of Students9118352074

Answer»

Marks	No. of Students
15	9
25	11
35	8
45	35
55	20
65	7
75	4

Highest value (H) = 75

Lowest value (L) = 15

Range = Highest value − Lowest value

or, R = 75 − 15= 60 marks

Coefficient of Range = \(\frac{H-L}{H+L}=\frac{75-15}{75+15}=\frac{60}{90}=0.66\)

Hence, range is 60 and coefficient of range is 0.66

Discussion

31429.	The weight (in kg) for 10 students are 53, 47, 60, 55, 71, 65, 61, 68, 63, 70. What is the range of the data?(a) 17 kg(b) 23 kg(c) 24kg(d) 18kg
Answer» Correct option is (c) 24kg

Discussion

31430.	The height of 11 men were 61, 64, 68, 69, 67, 68, 66, 70, 65, 67 and 72 inches. Calculate the range If the shortest man is omitted, what is the percentage change in the range?
Answer» 1. Range = L = S = 72 – 61 = 11 inches. 2. New Range (Shortest man is omitted) = L – S = 72 – 64 = 8 change in range =11 – 8 = 3 inches Percentage change in range = 3/11 x 100 = 27.2%

31430.

The height of 11 men were 61, 64, 68, 69, 67, 68, 66, 70, 65, 67 and 72 inches. Calculate the range If the shortest man is omitted, what is the percentage change in the range?

Answer»

1. Range = L = S = 72 – 61 = 11 inches.

2. New Range (Shortest man is omitted)

= L – S = 72 – 64 = 8

change in range =11 – 8 = 3 inches

Percentage change in range = 3/11 x 100 = 27.2%

Discussion

31431.	Find the sum of the series 101 + 99 + 97 + .... + 47.
Answer» In this case, we have to first find the number of terms. Here a = 101, l = T_n = 47, d = 99 – 101 = –2 ∴ 47 = 101 + (n – 1) × (–2), where n = number of terms ⇒ 47 = 101 – 2n +2 ⇒ 2n = 103 – 47 = 56 ⇒ n = 28 ∴ S_n = \(\frac{n}{2}\) (a + l) ⇒ S₂₈ = \(\frac{28}{2}\) (101 + 47) = 14 × 148 = 2072.

31431.

Find the sum of the series 101 + 99 + 97 + .... + 47.

Answer»

In this case, we have to first find the number of terms.

Here a = 101, l = T_n = 47, d = 99 – 101 = –2

∴ 47 = 101 + (n – 1) × (–2), where n = number of terms

⇒ 47 = 101 – 2n +2

⇒ 2n = 103 – 47 = 56 ⇒ n = 28

∴ S_n = \(\frac{n}{2}\) (a + l)

⇒ S₂₈ = \(\frac{28}{2}\) (101 + 47)

= 14 × 148

= 2072.

Discussion

31432.	In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.
Answer» In ΔQTR, ∠TRS is an exterior angle. ∠QTR + ∠TQR = ∠TRS ∠QTR = ∠TRS − ∠TQR (1) For ΔPQR, ∠PRS is an external angle. ∠QPR + ∠PQR = ∠PRS ∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors) ∠QPR = 2(∠TRS − ∠TQR) ∠QPR = 2∠QTR [By using equation (1)] ∠QTR = 1/2 ∠QPR

31432.

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

Answer»

In ΔQTR, ∠TRS is an exterior angle.

∠QTR + ∠TQR = ∠TRS

∠QTR = ∠TRS − ∠TQR (1)

For ΔPQR, ∠PRS is an external angle.

∠QPR + ∠PQR = ∠PRS

∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)

∠QPR = 2(∠TRS − ∠TQR)

∠QPR = 2∠QTR [By using equation (1)]

∠QTR = 1/2 ∠QPR

Discussion

31433.	In an examination, 70% of the candidates passed in English, 80% passed in Mathematics and 10% failed in both these subjects. If 144 candidates passed in both, then the total number of candidates was (a) 125 (b) 200 (c) 240 (d) 375
Answer» (c) 240 Let the total number of candidates be x. Given, 70% of x passed in English 80% of x passed in Maths 144 passed in English and Maths both 10% of x failed in English and Maths both \(\therefore\) 90% of x passed in English and Maths both. \(\therefore\) 90% of x = 70% x + 80% of x – 144 \(\Rightarrow\) 60% of 144 \(\Rightarrow\) x = \(\frac{144\times100}{60}\) = 240.

31433.

In an examination, 70% of the candidates passed in English, 80% passed in Mathematics and 10% failed in both these subjects. If 144 candidates passed in both, then the total number of candidates was (a) 125 (b) 200 (c) 240 (d) 375

Answer»

(c) 240

Let the total number of candidates be x.

Given, 70% of x passed in English

80% of x passed in Maths

144 passed in English and Maths both

10% of x failed in English and Maths both

\(\therefore\) 90% of x passed in English and Maths both.

\(\therefore\) 90% of x = 70% x + 80% of x – 144

\(\Rightarrow\) 60% of 144

\(\Rightarrow\) x = \(\frac{144\times100}{60}\) = 240.

Discussion

31434.	In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Answer» Given at start he has to run 24m to get the first potato then 28 m as the next potato is 4m away from first and so on Hence the sequence of his running will be 24, 28, 32 … There are 20 terms in sequence as there are 20 potatoes Hence only to get potatoes from starting point he has to run 24 + 28 + 32 + … up to 20 terms This is only from starting point to potato but he has to get the potato back to starting point hence the total distance will be twice that is Total distance ran = 2 × (24 + 28 + 32…) … 1 Let us find the sum using he formula to find sum of n terms of AP That is S_n= n/2 (2a + (n – 1) d) There are 20 terms hence n = 20 ⇒ S₂₀ = (20/2) (2 (24) + (20 – 1) 4) On simplification we get ⇒ S₂₀ = 10(48 + 19(4)) ⇒ S₂₀ = 10(48 + 76) On computing we get ⇒ S₂₀ = 10 × 124 ⇒ S₂₀ = 1240 m Using equation 1 ⇒ Total distance ran = 2 × 1240 ⇒ Total distance ran = 2480 m Hence total distance he has to run is 2480 m

31434.

In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Answer»

Given at start he has to run 24m to get the first potato then 28 m as the next potato is 4m away from first and so on

Hence the sequence of his running will be 24, 28, 32 …

There are 20 terms in sequence as there are 20 potatoes

Hence only to get potatoes from starting point he has to run

24 + 28 + 32 + … up to 20 terms

This is only from starting point to potato but he has to get the potato back to starting point hence the total distance will be twice that is

Total distance ran = 2 × (24 + 28 + 32…) … 1

Let us find the sum using he formula to find sum of n terms of AP

That is S_n= n/2 (2a + (n – 1) d)

There are 20 terms hence n = 20

⇒ S₂₀ = (20/2) (2 (24) + (20 – 1) 4)

On simplification we get

⇒ S₂₀ = 10(48 + 19(4))

⇒ S₂₀ = 10(48 + 76)

On computing we get

⇒ S₂₀ = 10 × 124

⇒ S₂₀ = 1240 m

Using equation 1

⇒ Total distance ran = 2 × 1240

⇒ Total distance ran = 2480 m

Hence total distance he has to run is 2480 m

Discussion

31435.	Solve:3 (x + 5) = 18
Answer» 3 (x + 5) = 18 => 3x + 15 = 18 => 3x = 18 -15 = 3 x = 3/3 = 1

31435.

Solve:3 (x + 5) = 18

Answer»

3 (x + 5) = 18

=> 3x + 15 = 18

=> 3x = 18 -15 = 3

x = 3/3 = 1

Discussion

31436.	Solve:7x -3x +2 = 22
Answer» 7x -3x +2 = 22 => 7x - 3x = 22 - 2 => 4x = 20 => x = 20/4 => x = 5

31436.

Solve:7x -3x +2 = 22

Answer»

7x -3x +2 = 22

=> 7x - 3x = 22 - 2

=> 4x = 20

=> x = 20/4

=> x = 5

Discussion

31437.	A number decreased by 8 equals 26, find the number.
Answer» Let required number = A ∴ According to the sum : x – 8 = 26 ⇒ A = 26 + 8 ⇒ A = 34 ∴ Required number = 34

31437.

A number decreased by 8 equals 26, find the number.

Answer»

Let required number = A

∴ According to the sum :

x – 8 = 26

⇒ A = 26 + 8

⇒ A = 34

∴ Required number = 34

Discussion

31438.	The sum of three consecutive numbers is 54. Taking the middle number as x, find:(i) expression for the smallest number and the largest number.(ii) the three numbers.
Answer» Sum of three consecutive numbers = 54 Middle number = x (i) The first number = x – 1 and third number = x + 1 (ii) ∴ x + x-1+x+1 = 54 ⇒ 3x = 54 ⇒ x= 54/3 = 18 ∴ First number =18-1 = 17 and third number =18 + 1 = 19 ∴ Three required numbers are 17, 18,19

31438.

The sum of three consecutive numbers is 54. Taking the middle number as x, find:(i) expression for the smallest number and the largest number.(ii) the three numbers.

Answer»

Sum of three consecutive numbers = 54

Middle number = x

(i) The first number = x – 1

and third number = x + 1

(ii) ∴ x + x-1+x+1 = 54

⇒ 3x = 54

⇒ x= 54/3 = 18

∴ First number =18-1 = 17

and third number =18 + 1 = 19

∴ Three required numbers are 17, 18,19

Discussion

31439.	Find three consecutive integers such that their sum is 78.
Answer» Sum of three consecutive numbers = 78 Let first number = x Then second number = x + 1 and third number = x + 2 Then x + x+1+x + 2 = 78 ⇒ 3x + 3 = 78 ⇒ 3x = 78 – 3 = 75 ⇒ x = 75/ 3 = 25 ∴ First number = 25 Second number = 25 + 1 = 26 and third number = 26 + 1 = 27 Then the three required numbers are 25, 26,27

31439.

Find three consecutive integers such that their sum is 78.

Answer»

Sum of three consecutive numbers = 78

Let first number = x

Then second number = x + 1

and third number = x + 2

Then x + x+1+x + 2 = 78

⇒ 3x + 3 = 78

⇒ 3x = 78 – 3 = 75

⇒ x = 75/ 3 = 25

∴ First number = 25

Second number = 25 + 1 = 26

and third number = 26 + 1 = 27

Then the three required numbers are 25, 26,27

Discussion

31440.	13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Can you find two other such pairs?
Answer» First pair is 12 and 21 Squares of numbers, 12² = 144 and 21² = 441 Second pair is 102 and 201 Squares of numbers, 102² = 10404 and 201² = 40401

31440.

13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Can you find two other such pairs?

Answer»

First pair is 12 and 21

Squares of numbers, 12² = 144 and 21² = 441

Second pair is 102 and 201

Squares of numbers, 102² = 10404 and 201² = 40401

Discussion

31441.	Floor of a room is of dimensions 5m × 4m and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find area of floor that remains uncovered with tiles. (use π = 3.14)
Answer» Length of the floor = l = 5 m Breadth of the floor = b = 4 m ∴ Area of the floor = l × b = 5 × 4 = 20 m² Now, Diameter of each circular tile = 50 cm ∴ Radius of each circular tile = r = 25 cm = 0.25 m Area of one circular tile = πr² = 3.14 × (0.25)² = 0.19625 m² Area of such 80 tiles = 80 × 0.19625 = 15.7 m² Area of the floor that remains uncovered with tiles = Area of the floor – Area of all 80 circular tiles = 20 – 15.7 = 4.3 m² Hence, the required area of floor that remains uncovered with tiles is 4.3 m².

31441.

Floor of a room is of dimensions 5m × 4m and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find area of floor that remains uncovered with tiles. (use π = 3.14)

Answer»

Length of the floor = l = 5 m

Breadth of the floor = b = 4 m

∴ Area of the floor = l × b = 5 × 4 = 20 m²

Now, Diameter of each circular tile = 50 cm

∴ Radius of each circular tile = r = 25 cm = 0.25 m

Area of one circular tile = πr²

= 3.14 × (0.25)²

= 0.19625 m²

Area of such 80 tiles = 80 × 0.19625

= 15.7 m²

Area of the floor that remains uncovered with tiles = Area of the floor – Area of all 80 circular tiles

= 20 – 15.7 = 4.3 m²

Hence, the required area of floor that remains uncovered with tiles is 4.3 m².

Discussion

31442.	A class room is 7m long, 6.5m wide and 4m high. It has one door 3m × 1.4m and three windows each measuring 2m × 1m The interior walls are to be color-washed. The contractor charges Rs. 15per sq. m. Find the cost of colour washing.
Answer» l = 7m, n = 6.5m and h = 4m ∴ Area of the room = 2( l + b)h = 2(7 + 6.5) 4 = 108m² Area of door = 3 × 1.4 = 4.2m² Area of one window = 3 × 2 = 6m² ∴ Area of 3 windows = 3 × 2 = 6m² ∴ Area of the walls of the room to be colour washed = 108 - (4.2 + 6) = 108 - 10.2 = 97.8m² ∴ Cost of colour washing @ Rs. 15 per square metre = Rs. 97.8 × 15 = Rs. 1467.

31442.

A class room is 7m long, 6.5m wide and 4m high. It has one door 3m × 1.4m and three windows each measuring 2m × 1m The interior walls are to be color-washed. The contractor charges Rs. 15per sq. m. Find the cost of colour washing.

Answer»

l = 7m, n = 6.5m and h = 4m

∴ Area of the room = 2( l + b)h = 2(7 + 6.5) 4 = 108m²

Area of door = 3 × 1.4 = 4.2m²

Area of one window = 3 × 2 = 6m²

∴ Area of 3 windows = 3 × 2 = 6m²

∴ Area of the walls of the room to be colour washed = 108 - (4.2 + 6)

= 108 - 10.2 = 97.8m²

∴ Cost of colour washing @ Rs. 15 per square metre = Rs. 97.8 × 15 = Rs. 1467.

Discussion

31443.	Can a normal convex lens behave like a concave lens and vice-versa?
Answer» Yes. If a lens is kept in a medium whose refractive index is more than that of the material of the lens, the normal convex lens starts behaving like a concave lens and vice-versa.

Discussion

31444.	The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm2 of the surface is 10 paise. Find the total cost to paint the vessel all over.
Answer» Given that, internal diameter of hollow hemisphere, r = 10.5 cm External radius, R = 12.6 cm Total surface area = 2πR² + 2πr² + π(R² - r²) = 2π(12.6)² + 2π(10.5)² + π((12.6)² – (10.5)²) = 1843.38 cm² Given that, cost of painting 1 cm²of surface = 10 paise Therefore, total cost to paint 1843.38 cm² = \(\frac{1843.38\times10}{100}\) Rs 184.34

31444.

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm2 of the surface is 10 paise. Find the total cost to paint the vessel all over.

Answer»

Given that,

internal diameter of hollow hemisphere, r = 10.5 cm

External radius, R = 12.6 cm

Total surface area = 2πR² + 2πr² + π(R² - r²)

= 2π(12.6)² + 2π(10.5)² + π((12.6)² – (10.5)²)

= 1843.38 cm²

Given that,

cost of painting 1 cm²of surface = 10 paise

Therefore, total cost to paint 1843.38 cm² = \(\frac{1843.38\times10}{100}\)

Rs 184.34

Discussion

31445.	The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kL of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.
Answer» Given: Internal diameter of hemispherical tank = 14 m Internal radius of hemispherical tank = 14/2 - 7 m Tank contains 50kl = 50m³ of water (since, 1kl = 1m³) Volume of a hemispherical tank = 2/3 πr³ Where π = 22/7 r = radius of hemispherical tank Thus, Volume of hemispherical tank = 2/3 x 22/7 x (7)³ = 2156/3 = 718.66 m³ Volume of water pumped into the tank = Volume of hemispherical tank – 50m³ Volume of water pumped into the tank = 718.66 – 50 = 668.66 m³

31445.

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kL of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

Answer»

Given: Internal diameter of hemispherical tank = 14 m

Internal radius of hemispherical tank = 14/2 - 7 m

Tank contains 50kl = 50m³ of water (since, 1kl = 1m³)

Volume of a hemispherical tank = 2/3 πr³

Where π = 22/7

r = radius of hemispherical tank

Thus,

Volume of hemispherical tank = 2/3 x 22/7 x (7)³

= 2156/3

= 718.66 m³

Volume of water pumped into the tank = Volume of hemispherical

tank – 50m³

Volume of water pumped into the tank = 718.66 – 50 = 668.66 m³

Discussion

31446.	Using factor theorem, factorize the polynomials:x4 – 7x3 + 9x2 + 7x - 10
Answer» Let f(x) = x⁴ – 7x³ + 9x² + 7x- 10 Constant term = -10 Factors of -10 are ±1, ±2, ±5, ±10 Let x – 1 = 0 or x = 1 f(1) = (1)⁴ – 7(1)³ + 9(1)² + 7(1) – 10 = 1 – 7 + 9 + 7 - 10 = 0 f(1) = 0 Let x + 1 = 0 or x = -1 f(-1) = (-1)⁴ – 7(-1)³ + 9(-1)² + 7(-1) – 10 = 1 + 7 + 9 – 7 - 10 = 0 f(-1) = 0 Let x – 2 = 0 or x = 2 f(2) = (2)⁴ – 7(2)³ + 9(2)² + 7(2) – 10 = 16 – 56 + 36 + 14 – 10 = 0 f(2) = 0 Let x – 5 = 0 or x = 5 f(5) = (5)⁴ – 7(5)³ + 9(5)² + 7(5) – 10 = 625 – 875 + 225 + 35 – 10 = 0 f(5) = 0 Therefore, (x – 1), (x + 1), (x – 2) and (x - 5) are factors of f(x) Hence f(x) = (x – 1) (x – 1) (x – 2) (x - 5).

31446.

Using factor theorem, factorize the polynomials:x4 – 7x3 + 9x2 + 7x - 10

Answer»

Let f(x) = x⁴ – 7x³ + 9x² + 7x- 10

Constant term = -10

Factors of -10 are ±1, ±2, ±5, ±10

Let x – 1 = 0 or x = 1

f(1) = (1)⁴ – 7(1)³ + 9(1)² + 7(1) – 10

= 1 – 7 + 9 + 7 - 10

= 0

f(1) = 0

Let x + 1 = 0 or x = -1

f(-1) = (-1)⁴ – 7(-1)³ + 9(-1)² + 7(-1) – 10

= 1 + 7 + 9 – 7 - 10

= 0

f(-1) = 0

Let x – 2 = 0 or x = 2

f(2) = (2)⁴ – 7(2)³ + 9(2)² + 7(2) – 10

= 16 – 56 + 36 + 14 – 10

= 0

f(2) = 0

Let x – 5 = 0 or x = 5

f(5) = (5)⁴ – 7(5)³ + 9(5)² + 7(5) – 10

= 625 – 875 + 225 + 35 – 10

= 0

f(5) = 0

Therefore, (x – 1), (x + 1), (x – 2) and (x - 5) are factors of f(x)

Hence f(x) = (x – 1) (x – 1) (x – 2) (x - 5).

Discussion

31447.	Using factor theorem, factorize each of the following polynomial:x3 + 13x2 + 32x + 20
Answer» Let, f (x) = x³ + 13x² + 32x + 20 The factors of the constant term + 20 are \(\pm\) 1, \(\pm\) 2,\(\pm\) 4, \(\pm\) 5, \(\pm\) 10 and 20 Putting x = -1, we have f (-1) = (-1)³ + 13 (-1)² + 32 (-1) + 20 = -1 + 13 – 32 + 20 = 0 So, (x + 1) is a factor of f (x) Let us now divide f (x) = x³ + 13x² + 32x + 20 by (x + 1) to get the other factors of f (x) Using long division method, we get x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) x² + 2x + 20 = x² + 10x + 2x + 20 = x (x + 10) + 2 (x + 10) = (x + 10) (x + 2) Hence, x³ + 13x² + 32x + 20 = (x + 1) (x + 10) (x + 2)

31447.

Using factor theorem, factorize each of the following polynomial:x3 + 13x2 + 32x + 20

Answer»

Let, f (x) = x³ + 13x² + 32x + 20

The factors of the constant term + 20 are \(\pm\) 1, \(\pm\) 2,\(\pm\) 4, \(\pm\) 5, \(\pm\) 10 and 20

Putting x = -1, we have

f (-1) = (-1)³ + 13 (-1)² + 32 (-1) + 20

= -1 + 13 – 32 + 20

= 0

So,

(x + 1) is a factor of f (x)

Let us now divide

f (x) = x³ + 13x² + 32x + 20 by (x + 1) to get the other factors of f (x)

Using long division method, we get

x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20)

x² + 2x + 20 = x² + 10x + 2x + 20

= x (x + 10) + 2 (x + 10)

= (x + 10) (x + 2)

Hence,

x³ + 13x² + 32x + 20 = (x + 1) (x + 10) (x + 2)

Discussion

31448.	Find the greatest common factor (GCF/HCF) of the given polynomials: 9x2, 15x2y3, 6xy2 and 21x2y2
Answer» We know that the numerical coefficients of given numerical are 9, 15, 16 and 21 Greatest common factor of 9, 15, 16 and 21 is 3. Common literals appearing in given numerical are x and y Smallest power of x in four monomials is 1 Smallest power of y in four monomials is 0 Monomials of common literals with smallest power is x ∴ The greatest common factor = 3x

31448.

Find the greatest common factor (GCF/HCF) of the given polynomials: 9x2, 15x2y3, 6xy2 and 21x2y2

Answer»

We know that the numerical coefficients of given numerical are 9, 15, 16 and 21

Greatest common factor of 9, 15, 16 and 21 is 3.

Common literals appearing in given numerical are x and y

Smallest power of x in four monomials is 1

Smallest power of y in four monomials is 0

Monomials of common literals with smallest power is x

∴ The greatest common factor = 3x

Discussion

31449.

Using factor theorem, factorize the polynomials:2x4 – 7x3 – 13x2 + 63x – 45

Answer»

Let f(x) = 2x⁴ – 7x³ – 13x² + 63x – 45

Constant term = -45

Factors of -45 are ±1, ±3, ±5, ±9, ±15, ±45

Here coefficient of x⁴ is 2.

So possible rational roots of f(x) are ±1, ±3, ±5, ±9, ±15, ±45, ±1/2,±3/2,±5/2,±9/2,±15/2,±45/2

Let x – 1 = 0 or x = 1

f(1) = 2(1)⁴ – 7(1)³ – 13(1)² + 63(1) – 45

= 2 – 7 – 13 + 63 – 45

= 0

f(1) = 0

f(x) can be written as,

f(x) = (x - 1) (2x³ – 5x² -18x + 45)

or f(x) = (x - 1) g(x) …(1)

Let x – 3 = 0 or x = 3

f(3) = 2(3)⁴ – 7(3)³ – 13(3)² + 63(3) – 45

= 162 – 189 – 117 + 189 – 45

= 0

f(3) = 0

Now, we are available with 2 factors of f(x), (x – 1) and (x – 3)

Here g(x) = 2x² (x-3) + x(x-3) -15(x-3)

Taking (x-3) as common

= (x-3)(2x² + x – 15)

= (x-3)(2x²+6x – 5x -15)

= (x-3)(2x-5)(x+3)

= (x-3)(x+3)(2x-5) ….(2)

From (1) and (2)

f(x) =(x-1) (x-3)(x+3)(2x-5)

Discussion

31450.

Factorize each of the following polynomials:(i) x3 + 13x2 + 31x - 45 given that x+9 is a factor(ii) 4x3 + 20x2 + 33x + 18 given that 2x+3 is a factor.

Answer»

(i) Let, f (x) = x³ + 13x² + 31x - 45

Given that (x + 9) is a factor of f (x)

Let us divide f (x) by (x + 9) to get the other factors

By using long division method, we have

f (x) = x³ + 13x² + 31x - 45

= (x + 9) (x² + 4x – 5)

Now,

x² + 4x – 5 = x² + 5x – x – 5

= x (x + 5) – 1 (x + 5)

= (x – 1) (x + 5)

f (x) = (x + 9) (x + 5) (x – 1)

Therefore,

x³ + 13x² + 31x - 45 = (x + 9) (x + 5) (x – 1)

(ii) Let, f (x) = 4x³+20x²+33x+18

Given that (2x + 3) is a factor of f (x)

Let us divide f (x) by (2x + 3) to get the other factors

By long division method, we have

4x³ + 20x² + 33x + 18 = (2x + 3) (2x² + 7x + 6)

2x² + 7x + 6 = 2x² + 4x + 3x + 6

= 2x (x + 2) + 3 (x + 2)

= (2x + 3) (x + 2

4x³ + 20x² + 33x + 18 = (2x + 3) (2x + 3) (x + 2)

= (2x + 3)2 (x + 2)

Hence,

4x³ + 20x² + 33x + 18 = (2x + 3)² (x + 2)

Discussion

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If variance (σ) = √(∑fd2/k - (∑fd/30)2) then value of k is –(A) 10(B) 20(C) 30(D) 60

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From the following data calculate range and coefficient of range:Marks15253545556575No. of Students9118352074

The weight (in kg) for 10 students are 53, 47, 60, 55, 71, 65, 61, 68, 63, 70. What is the range of the data?(a) 17 kg(b) 23 kg(c) 24kg(d) 18kg

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Find the sum of the series 101 + 99 + 97 + .... + 47.

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

In an examination, 70% of the candidates passed in English, 80% passed in Mathematics and 10% failed in both these subjects. If 144 candidates passed in both, then the total number of candidates was (a) 125 (b) 200 (c) 240 (d) 375

In a potato race 20 potatoes are placed in a line at intervals of 4 metres with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?

Solve:3 (x + 5) = 18

Solve:7x -3x +2 = 22

A number decreased by 8 equals 26, find the number.

The sum of three consecutive numbers is 54. Taking the middle number as x, find:(i) expression for the smallest number and the largest number.(ii) the three numbers.

Find three consecutive integers such that their sum is 78.

13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Can you find two other such pairs?

Floor of a room is of dimensions 5m × 4m and it is covered with circular tiles of diameters 50 cm each as shown in figure. Find area of floor that remains uncovered with tiles. (use π = 3.14)

A class room is 7m long, 6.5m wide and 4m high. It has one door 3m × 1.4m and three windows each measuring 2m × 1m The interior walls are to be color-washed. The contractor charges Rs. 15per sq. m. Find the cost of colour washing.

Can a normal convex lens behave like a concave lens and vice-versa?

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm2 of the surface is 10 paise. Find the total cost to paint the vessel all over.

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kL of water. Water is pumped into the tank to fill to its capacity. Calculate the volume of water pumped into the tank.

Using factor theorem, factorize the polynomials:x4 – 7x3 + 9x2 + 7x - 10

Using factor theorem, factorize each of the following polynomial:x3 + 13x2 + 32x + 20

Find the greatest common factor (GCF/HCF) of the given polynomials: 9x2, 15x2y3, 6xy2 and 21x2y2

Using factor theorem, factorize the polynomials:2x4 – 7x3 – 13x2 + 63x – 45

Factorize each of the following polynomials:(i) x3 + 13x2 + 31x - 45 given that x+9 is a factor(ii) 4x3 + 20x2 + 33x + 18 given that 2x+3 is a factor.

यदि किसी आवृत्ति वितरण में (Q3 – Q2) < (Q2 – Q1) हो, तो आवृत्ति वितरण का प्रकार बताइए ।

Which of the following statements is true for a symmetric distribution?(a) Q3 = 2M – Q1(b) Q2 – Q3 = Q2 – Q1(c) Q3 + Q1 > 2M(d) Q3 + Q1 < 2M

संमित आवृत्ति वितरण के लिए कौन-सा विधान सत्य है ? (A) Q3 = 2M – Q1(B) Q2 – Q3 = Q2 – Q1(C) Q3 + Q1 > 2M(D) Q3 + Q1 < 2M