This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 301. |
Find the values of (-2) × 5 = __________ = -(2 × 5) = – 10 |
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Answer» The values is 2 × (-5) |
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| 302. |
What happens when we multiply a negative integer with a positive integer? |
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Answer» When a negative integer is multiplied with a positive integer then negative integer is found, it can be verified from following examples – (i) – 1 x 4 = – 4 = 0 – 4 |
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| 303. |
Find the following products:(i) (-12) x (-15)(ii) (-25) x (-4)(iii) (-17) x (-11) |
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Answer» (i) (-12) x (-15) = 180 |
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| 304. |
The sum of two integers is 30. If one of the integers is –42, then find the other. |
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Answer» The sum of two integers is 30. If one of the integers is –42, then the other is 72. |
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| 305. |
Are (-42) × (-7) and (-7) × (-42) equal ? Which is this law ? |
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Answer» (-42) × (-7) = + 294 (-7) × (-42) = + 294 ∴ (- 42) × (- 7) = (- 7) × (- 42) It is multiplicative commutative law. |
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| 306. |
Find the values of (-6) × 7 = __________ = -(6 × 7) = – 42 |
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Answer» The values is 6 × (-7) |
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| 307. |
fill in the blanks to make the statements true._______× (–93) = 93 |
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Answer» -1× (–93) = 93 |
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| 308. |
Fill in the blanks using <, = or > :–60 _______ 50 |
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Answer» Fill in the blanks using <, = or > : –60 < 50 |
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| 309. |
Calculate : (- 12) ÷ (+ 6) |
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Answer» (- 12) ÷ ( + 6) We know, (- a) ÷ (- b) = – (a ÷ b) = (- 12) ÷ 6 = – 2 |
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| 310. |
Start with (- 1) x 5 and make pattern to show that (- 1) x (- 1) = + 1. |
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Answer» Pattern is : (-1) x 5 = -5 (-1) x 2 = -2 |
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| 311. |
Identify True/False. Correct the false statements and write :(i) Multiplication of integers is closed.(ii) Division of integers is closed.(iii) Division of integers is not commutative but multiplication is commutative.(iv) Multiplication of integers, is distributive over addition.(v) Division of integers is distributive on subtraction. |
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Answer» (i) True |
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| 312. |
Fill in the blanks using <, = or > :0 _______ 1 |
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Answer» Fill in the blanks using <, = or > : 0 < 1 |
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| 313. |
Fill in the blanks using <, = or >.(-2) + (-5) + (-6) ___ (-3) + (-4) + (-6) |
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Answer» = (-2) + (-5) + (-6) = -2 – 5 – 6 = -13 (-3) + (-4) + (-6) = -3 – 4 – 6—13 And-13 = -13 |
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| 314. |
Value of (-3) x 5 will be(A) – 15(B) 15(C) 30(D) – 30 |
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Answer» (A) – 15 (-3) x 5 = -15 |
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| 315. |
Fill in the blanks using :(–71) + (+9) _______ (–81) + (–9) |
| Answer» (–71) + (+9) > (–81) + (–9) | |
| 316. |
Fill in the blanks using <, = or >.0 ___ -2 |
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Answer» 0 > -2 0 > -2 |
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| 317. |
On the number line, –15 is to the _______ of zero. |
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Answer» On the number line, –15 is to the left of zero. |
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| 318. |
In the figure below, the square ABCD is made up of two parts, X and Y.The part X, is formed by a semicircle and line AB. The perimeter of X is 36 cm and perimeter of the shaded part Y is 64 cm.a. Find the perimeter of square ABCDb. Find the area of shaded region Y |
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Answer» `a+pir=36cm-(1)` `pir+3a=64-(2)` subtracting equation 2 from equation 1 3a-a=64-36 a=14cm `a+pir=36cm` `pir=22cm` area of semicircle =`1/2*pir^2=22/2*22/22*7=77cm^2` perimeter of square ABCD =4a=4*14=56cm area of y=area of square- area of semicircle`=14*14-77` `=119cm^2`. |
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| 319. |
Mrs. Kapoor has certain sum of money to buy fruits. She can buy `n` mangoes at Rs.1.60 each and have Rs 0.80 left. Alternatively, she can buy `(n+10)` apples at Rs0.70 each and have Rs0.10 left. If Mrs. Kapoor buys 3 mangoes and uses the rest of the money to buy apples, then |
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Answer» Let Mrs. Kapoor has `x` money in rupees. Then, `1.60n+0.8 = x->(1)` `0.7(n+10)+0.10 = x->(2)` From (1) and (2), `1.6n+0.8 = 0.7n+7+0.10` `=>0.9n = 6.3=> n = 7` Putting value of `n` in (1), `1.6(7)+0.8 = x` `=> x = 11.2+0.8 => x = 12` So, Mrs. Kapoor has `12` rupees. |
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| 320. |
Write the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 in this order and insert ‘+ ‘or ‘–’ between them to get the result (a) 5 (b) –3 |
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Answer» (a) 0 + 1 – 2 + 3 – 4 + 5 – 6 + 7 –8 + 9 = 5 (b) 0 – 1 – 2 + 3 + 4 – 5 + 6 – 7 + 8 – 9 = –3 |
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| 321. |
A boy has Rs 728 in denomination of Rs 2 notes,Rs 5 notes and Rs 10 notes in the ratio of 3:2:4 respectively.What is the total number of notes he has? |
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Answer» Let the boy has `3x` 2 rupees notes. Then, he will have `2x` Rs 5 notes and `4x` Rs 10 notes. Total number of notes `= 3x+2x+4x = 9x` Also, we are given, `2(3x)+5(2x)+10(4x) = 728` `=>6x+10x+40x =728=> 56x = 728` `=> x = 13` So, he will have total number of notes `= 9**13 = 117` |
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| 322. |
A sum of rs 800 is in the form of denomination of rs 10 and rs 20.If the total number of notes are 50, find the number of notes of each type. |
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Answer» let Rs 10=x Let Rs21=y x+y=50-(1) 10x+120y=800 x+2y=80-(2) subtracting equation 1 from equation 2 y=30 x=20 20 notes of 10 Rs 30 notes of 20 Rs. |
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| 323. |
A rectangular lawn 70 m by 50 m has two roads, each 5 m wide, running through its middle, one parallel to its length and other parallel to its breadth. Find the cost of constructing the roads at Rs 120 per `m^2` |
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Answer» Area of the road parallel to the length of the rectangular lawn ` = 70**5 = 350m^2` Area of the road parallel to the length of the rectangular lawn ` = 50**5 = 250m^2` There is a common area between these two roads that is a square of `5m` side. `:.` Area of common area between two roads `= 5**5=25m^2` `:.` Area of both roads ` = 350+250-25 = 575m^2` `:.` Cost of constructing roads at Rs`120` per `m^2 = 575**120 = 69000` Rs. |
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| 324. |
Compute each of the following: (a) 30 + (–25) + (–10) (b) (–20) + (–5) (c) 70 + (–20) + (–30) (d) –50 + (–60) + 50 (e) 1 + (–2) + (– 3) + (– 4) (f) 0 + (– 5) + (– 2) (g) 0 – (–6) – (+6) (h) 0 – 2 – (–2) |
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Answer» (a) –5 (b) – 25 (c) 20 (d) – 60 (e) – 8 (f) – 7 (g) 0 (h) 0 |
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| 325. |
Factorise: `p^2q-pr^2-pq+r^2` |
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Answer» `p^2q-pq-pr^2+r^2` `pq(p-1)-r^2(p-1)` `(pq-r^2)(p-1)`. |
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| 326. |
If `x/5 , x , x/4 , x/2 and x/3` are some observations, where x > 0. If their median is 10 then find the mean. |
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Answer» Median=`((5+1)/2)^(th) term` `x/3=10,x=30` Mean=(sum of terms/5)=`(30+15+10+7.5+6)/5=13.7`. |
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| 327. |
ABC is an equilateral triangle, and AD and BE are perpendiculars to BC and AC respectively. Prove that:1. AD=BE2. BD=CE`[ DeltaABD=DeltaBCE.]` |
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Answer» We know perpendicular in an equilateral triangle are bisectors of opposite sides. `:. BD= DC = (BC)/2->(1)` `AE = CE = (AC)/2->(2)` As `AC = BC` `:. (BC)/2 =(AC)/2` `:. BD = CE` Now, in `Delta ABD` and `Delta BCE` `AB =BC` `BD = CE` `/_ADB =/_BEC=90^@` So, by `R.H.S. ` congruency rule, `Delta ABD ~= Delta BCE` `:. AD = BE` |
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| 328. |
7 – {7 – (7 – \(\overline{7-1}\) )} = ……………? A) 0 B) 7 C) 6 D) 1 |
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Answer» Correct option is (D) 1 7- {7 - (7 - \(\overline {7-1}\))} = 7[7 - {7 - ( 7 -1)}] = 7 - {7 - (7 - 6)} = 7 - (7 - 1) = 7 - 6 = 1 Correct option is D) 1 |
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| 329. |
(- 1) × 2 × (- 3) × 4 × (- 5) = ……….. ? A) 60 B) 120 C) -120 D) – 240 |
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Answer» Correct option is C) -120 -2 x -3 x -20 = 6 x -20 = -120 |
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| 330. |
(- 576) × 1153 × 405 × 0 × 11 = ………..? A) 0 B) 11 C) – 50730 D) None of these |
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Answer» Correct option is A) 0 if any number multiplied with 0 then product equals to zero . ∴ (-576) x 1153 x 405 x 0 x 11 = 0 |
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| 331. |
Put appropriate symbol > or < in the boxes given.i) -1 ⬜ 0 ii) -3 ⬜ -7 iii) -10 ⬜ +10 |
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Answer» i) -1 < 0 ii) -3 > -7 iii) -10 < +10 |
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| 332. |
Write True or False against each of the following statements. i) -7 is on the right side of -6 on the number line. ii) Zero is a positive number. iii) 29 is on the right side of zero on the number line. iv) -1 lies between the integers -2 and 1. v) There are nine integers between -5 and +5. |
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Answer» i) False ii) False iii) True iv) True v) True |
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| 333. |
State whether the statement are True or False.(–5) × (–7) is same as (–7) × (–5) |
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Answer» True (–5) × (–7) is same as (–7) × (–5). |
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| 334. |
Observe the following number line and answer the following questions.i) Which is the nearest positive integer to -1? ii) How many negative numbers you will find on the left side of Zero? iii) How many integers are there in between -3 and 7? iv) Write 3 integers lesser than -2. v) Write 3 integers more than -2. |
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Answer» i) +1 is the nearest positive integer to -1. ii) On the given number line negative numbers to left of zero are -1, -2, -3, -4, -5. So, there are 5 in number. iii) On the number line integers between -3 and 7 are -2, -1, 0, 1, 2, 3, 4, 5, 6. So, there are 9 in number. iv) Integers less than -2 means numbers left side of -2. They are -3, -4, -5, -6, -7, ……… So, 3 integers lesser than -2 are -3, -4, -5. v) Integers more than -2 means numbers right side of -2. They are -1, 0, 1, 2, 3, 4, …….. So, 3 integers more than -2 are -1, 0, 1. (i) zero(ii) 5 (iii) 10 (iv) -3,-4,-5 (v) -1,0,1 |
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| 335. |
(- 289) ÷ (- 17) = …………….A) 1/17B) – 17 C) 17 D) -1/17 |
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Answer» Correct option is C) 17 \(\frac{-17 \times17}{-17}\) = \(\frac{-17}{-1}\) = 17 |
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| 336. |
The temperature of particular day of a village Lambasingi is – 16°C. it is reduced to – 11°C on the next day, what is the present temperature?A) 27° B) – 27° C) – 5° D) 5° |
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Answer» Correct option is C) – 5° present temperature = -16° c - (-11° c) = -16° c + 11° c = (-16 + 11)° c = - 5° c |
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| 337. |
State whether the statement are True or False.(– 80) ÷ (4) is not same as 80 ÷ (–4) |
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Answer» False (– 80) ÷ (4) is not same as 80 ÷ (–4) |
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| 338. |
Find the odd one out* of the four options in the following: (a) (–2, 24) (b) (–3, 10) (c) (–4, 12) (d) (–6, 8) |
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Answer» Here – 2 × 24 = – 48, – 4 × 12 = – 48 and – 6 × 8 = – 48 All the pairs i.e. (–2, 24); (–4, 12); (–6, 8) give same answer on multiplication, whereas –3 × 10 = –30, gives a different answer. So, odd one is (b). |
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| 339. |
(-1)odd number + (-1)even number = ……………..?A) 1 B) 0 C) -1 D) 2 |
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Answer» Correct option is B) 0 ∵ (-1) odd number = -1 and (-1) even number = 1 ∴ (-1) odd number + (-1) even number = -1 + 1 = 0 |
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| 340. |
Find the odd one out of the four options given below:(a) (–3, –6) (b) (+1, –10) (c) (–2, –7) (d) (–4, –9) |
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Answer» Here –3 + (–6) = –9, +1 + (–10) = –9 and –2 + (–7) = –9 All the above pairs i.e. (–3, –6); (+1, –10); (–2, –7) give same answer on adding, whereas – 4 + (–9) = –13, gives a different answer. So, odd one out is (d). |
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| 341. |
Write a pair of integers whose sum is zero (0) but difference is 10. |
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Answer» Since sum of two integers is zero, one integer is the additive inverse of other integer, like – 3, 3; – 4, 4 etc. But the difference has to be 10. So, the integers are 5 and – 5 as 5–(–5) is 10. |
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| 342. |
State whether the given statement are true (T) or false (F) :The sum of any two negative integers is always greater than both the integers. |
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Answer» False Since, the sum of any two negative integers is always smaller than both the integers. |
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| 343. |
Write a pair of integers whose product is – 15 and whose difference is 8. |
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Answer» There are few pairs of integers whose product is – 15. e.g. – 1 × 15 – 3 × 5 3 × (– 5) 15 × (– 1) but difference of –3 and 5 or –5 and 3 is 8. So the required pair of integers is – 3, 5 and – 5, 3. |
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| 344. |
State whether the statements are True or False.Sum of two negative integers always gives a number smaller than both the integers. |
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Answer» Sum of two negative integers always gives a number smaller than both the integers. True |
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| 345. |
Write two integers which are smaller than –3, but their difference is greater than –3. |
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Answer» – 5 and – 4 are smaller than – 3 but their difference is (–4) – (–5) = 1 which is greater than – 3. or – 6 and –10 are smaller than – 3 but their difference is (–6) – (–10) = 4 which is greater than – 3. |
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| 346. |
Write four negative integers greater than -20 |
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Answer» -19, -18, -17, -16 |
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| 347. |
Find the sum of the pairs of integers: (a) – 6, – 4 (b) +3, – 4 (c) +4, –2 |
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Answer» (a) – 6 and – 4 both have negative signs. So, – 6 + (– 4) = – (6 + 4) = –10 (b) + 3 and – 4 have opposite signs. As 4 – 3 = 1, therefore + 3 + (– 4) = –1 (c) + 4 and –2 have opposite signs. So, 4 + (–2) = 4 – 2 = 2 |
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| 348. |
Subtract: (i) 3 from –4 (ii) –3 from –4 |
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Answer» (a) The additive inverse of 3 is –3. So, – 4 – 3 = – 4 + (–3) = – (4 + 3) = –7 (b) The additive inverse of –3 is + 3. So, – 4 – (–3) = – 4 + (+3) = –1 |
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| 349. |
An integer with positive sign (+) is always greater than(A) 0 (B) 1 (C) 2 (D) 3 |
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Answer» (A) 0 Positive integers are always coming right to the 0, so positive integers always greater than 0. |
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| 350. |
An integer with positive sign (+) is always greater than (A) 0 (B) 1 (C) 2 (D) 3 |
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Answer» The correct option is (A) 0. |
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