This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Who lives in the cow sheds?(a) Camel (b) Cow (c) Horse (d) Goat |
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Answer» Cow lives in the cow sheds |
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| 2. |
Posing a problem Shyla made the graph as follows. Use the graph to write a problem involving fractions. How does Shyla spend her day? |
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Answer» Correct answer is Sleep→8hrs, Study→3hrs, Meals→2hrs, School→7hrs and Personal time 4 hrs. |
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| 3. |
Give suitable reasons A tarnished copper vessel regains its shine when rubbed with lemon. |
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Answer» Copper vessels tarnish due to the formation of basic copper carbonate which gets neutralized when rubbed with lemon and the copper vessel regains its shine. |
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| 4. |
For a kite to have balance while flying, each left and right sides of the horizontal support must be 2/3 as long as the bottom of the vertical support. Also, the top of vertical support must be 1/3 the bottom portion.(a) Suppose the bottom portion of the vertical support is 21 cm.(b) Length of upper vertical support = 1/3 × 21 cm = 7 cmThat is, OA = 7 cm(c) Total length of horizontal support is2 x 2/3 × 21 cm = 28 cmThat is, BD = 28 cmAnswer the following.1. If length of bottom of vertical support is 15 cm find the length of horizontal support.2. If total length of vertical support is 20 cm then find the length of each (left and right) side of horizontal support.3. If length of horizontal support is 12 cm, find the length of upper and lower vertical support. |
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Answer» Correct answer is 1. 20 cm 2. 40/9 cm 3. Length of bottom of vertical support = 9 cm Length of upper of vertical support = 3 cm |
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| 5. |
Meteorology: One measure of average global temperature shows how each year varies from a base measure. The table shows results for several years.Year19581964196519782002Difference from Base0.10°C–0.17°C–0.10°C(1/50)°c 0.54°CSee the table and answer the following:(a) Order the five years from coldest to warmest.(b) In 1946, the average temperature varied by –0.030C from the base measure. Between which two years should 1946 fall when the years are ordered from coldest to warmest? |
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Answer» (i) 1964, 1965, 1978, 1958, 2002 (ii) 1946 should fall between 1965 and 1978 |
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| 6. |
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers? |
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Answer» We have, Number of players = 17 Number of players to be select for team =11 Number of bowlers = 5 Number of bowlers to be select = 4 Number of players other than bowlers =12 4 bowlers out of 5 can be selected in 5C4 ways and 7(i.e., 11-4 = 7) other players can be selected out of 12 (i.e., 17 – 5 = 12) in 12C7 ways. ∴ Required number of ways = 5C4 x 12C7 = 5C1 x 12C5 ∴ nCr = nCn - r = 5((12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)) = 3960 |
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| 7. |
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king? |
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Answer» We have to select one king out of 4 kings and four other cards out of 48. This can be done in 4C, x 48C4 ways = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3)/(2 x 1) = 907200 ∴ Required number of ways = 4C1 x 48C4 = 4 x 2 x 47 x 46 x 45 = 778320 |
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| 8. |
If the selling price of 10 pens is the same as the cost price of 12 pens then gain percent is2%12%20%25% |
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Answer» (3) 20% let us consider the CP as x SP of 1 pen = x/10 CP of 1 pen = x/12 Since SP is more than CP its Gain Gain = SP – CP = x/10 – x/12 = x/60 ∴ Gain% = (Gain × 100) /CP = ((x/60) × 100) / (x/12) = 20% |
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| 9. |
On selling 100 pencils a man gains the selling price of 20 pencils. His gain percent is20%25%22 ½ %16 2/3 % |
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Answer» (2) 25% let us consider the CP of a pencil be x SP of 100 pencils = 100x Gain of 20 pencils = 20x ∴ we calculate CP = SP – Gain = 100x – 20x = 80x ∴ Gain% = (Gain × 100) /CP = ((20x) × 100) / 80x = 25% |
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| 10. |
Ravi buys some toffees at 5 for a rupee and sells them at 2 for a rupee. His gain percent is30%40%50%150% |
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Answer» (4) 150% we know that the CP of 1 toffee = ₹1/5 SP of 1 toffee = ₹1/2 Since SP is more than CP it’s a Gain Gain = SP – CP = ½ – 1/5 = 3/10 ∴ Gain% = (Gain × 100) /CP = ((3/10) × 100) / (1/5) = 150% |
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| 11. |
Ravi buys some toffees at 5 for a rupee and sells them at 2 for a rupee. His gain per cent is A. 30% B. 40% C. 50% D. 150% |
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Answer» Cost Price of 1 toffee = Rs.1/5 Selling Price of 1 toffee = Rs.1/2 Gain = SP – CP = 1/2 - 1/5 = 3/10 \(Gain\%=\frac{Gain\,\times\,100}{CP}\) \(=\frac{\frac{3}{10}\times100}{\frac{1}{5}}\) = 150% |
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| 12. |
Oranges are bought at 5 for Rs. 10 and sold at 6 for Rs.15. His gain per cent is A. 50% B. 40% C. 35% D. 25% |
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Answer» Cost Price of 1 Orange = Rs.10/5 = Rs.2 Selling Price of 1 Orange = Rs.15/6 = Rs.2.5 Gain = SP – CP = 2.5 - 2 = 0.5 Gain Percent = \(Gain\%=\frac{Gain\,\times\,100}{CP}\) = (0.5 × 100) / 2 = 25% |
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| 13. |
The marked price of an article is 10% more than the cost price and a discount of 10% is given on the marked price. The seller has A. no gain and no loss B. 1% gain C. 1% loss D. none of these |
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Answer» When two similar items are sold at same price, one at a gain and other at a loss of same percent. Then always a loss will be occurred. Loss % = (Common Loss and Gain Percent / 10)2 = (10/10)2 = (1)2 = 1 So, Loss will be 1%. |
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| 14. |
On selling a bat for ₹ 100, a man gains ₹20. His gain % is(a) 20% (b) 25% (c) 18% (d) 22% |
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Answer» (b) 25% Because, Selling price of bat = ₹ 100 Amount gain by selling bat = ₹20 Cost price of the bat = (100 – 20) = ₹ 80 Gain % = {(gain/CP) × 100} = {(20/80) × 100} = {(20/20) × 25} = 25% |
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| 15. |
By selling a radio for Rs. 950, a man loses 5%. What per cent shall he gain by selling it for Rs. 1040? A. 4% B. 4.5% C. 5% D. 9% |
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Answer» SP = Rs.950 Loss % = 5 CP = \(\frac{100}{100\,-\,Loss\%}\times SP\) = \(\frac{100}{100\,-\,5}\times 950\) = Rs.1000 New SP will be Rs.1040 Gain = SP - CP = 1040 – 1000 = Rs.40 \(Gain\%=\frac{Gain\,\times\,100}{CP}\) = (40 × 100) / 1000 = 4% |
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| 16. |
Prabha bought an article Rs. 374. Which included a discount of 15% on the market price and a sales tax of 10% on the reduced price. Find the market price of the article. |
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Answer» Let the market price of the article be Rs. x ∴ Discount 15% of x = 15/100 x X = 3x/20 ⇒ Remaining cost = x – 3x/20 = 17x/20 Sales tax = 10% ∴ price paid by prabha = 110/100 x 17x/20 = 187x/200 Given 187x/200 = 374 = x = 374 = 2000/187 = 400 ∴ Market price = Rs. 400 |
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| 17. |
On selling a jug for ₹ 144, a man loses (1/7) of his outlay. If it is sold for ₹ 189, what is the gain %?(a) 12.5% (b)25% (c) 30% (d)50% |
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Answer» (a) 12.5% Because, Let the CP be, ₹ x Then, x – (1/7)x = 144 = (7x –x) = (144 × 7) = x = (144 ×7)/6 = x = 168 ∴ CP = ₹ 168, New SP = ₹189 Gain = SP –CP = 189 – 168 = 21 Gain % = {(gain/CP) × 100} = {(21/168) × 100} = {2100/168) = 12.5% |
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| 18. |
On selling a chair forRs. 720, a man loses 25%. To gain 25% it must be sold for A. Rs.900 B. Rs.1200 C. Rs.1080 D. Rs.1440 |
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Answer» SP = Rs.720 Loss % = 25 CP \(=\frac{100}{100\,-\,Loss\%}\times SP\) \(=\frac{100}{100\,-\,25}\times 720\) = Rs.960 SP = \(\frac{100\,+\,Gain\%}{100}\times CP\) = \(\frac{100\,+\,25}{100}\times 960\) = Rs.1200 |
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| 19. |
By selling an umbrella for Rs.336, a shopkeeper loses 4%. At what price must he sell it to gain 4%? |
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Answer» Let x be the CP of an umbrella SP = \(\frac{100\,-\,Loss\%}{100}\times CP\) \(336=\frac{100\,-\,4}{100}\times x\) \(336=\frac{96x}{100}\) = Rs.350 So, CP of an umbrella is Rs.350. New SP to gain 4% SP = \(\frac{100\,+\,Gain\%}{100}\times CP\) \(=\frac{100\,+\,4}{100}\times350\) \(=\frac{104}{100}\times350\) = Rs.364 So, to gain 4% on Umbrella new Selling Price will be Rs.364. |
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| 20. |
A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of a 40% acid solution? |
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Answer» Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively. As per the question 50% of x + 25% of y = 40% of 10 ⇒ 0.50x + 0.25y = 4 ⇒ 2x + y = 16 ………(i) Since, the total volume is 10 liters, so x + y = 10 Subtracting (ii) from (i), we get x = 6 Now, putting x = 6 in (ii), we have 6 + y = 10 ⇒ y = 4 Hence, volume of 50% acid solution = 6litres and volume of 25% acid solution = 4litres. |
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| 21. |
If ten students are asked to measure the length of a piece of cloth upto a mm, using a metre scale, do you think their answers will be identical? Give reasons. |
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Answer» Answers of the students are likely to be different. Length of cloth needs to be measured up to a millimetre (mm) length. Hence, to obtain accurate and precise reading one must use measuring instrument having least count smaller than 1 mm. But least count of metre scale is 1 mm. As a result, even smallest uncertainty in reading would vary reading significantly. Also, skill of students doing measurement may also introduce uncertainty in observation. Hence, their answers are likely to be different. |
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| 22. |
Which are the even prime numbers? |
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Answer» The even prime numbers 2 . |
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| 23. |
The LCM and HCF of two positive numbers are 175 and 5 respectively. If the sum of the numbers is 60, what is the difference between them ? |
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Answer» Let the two numbers be 5a and 5b as HCF of the two numbers = 5 ∴ Product of the two number = HCF × LCM ⇒ 5a × 5b = 5 × 175 ⇒ ab = \(\frac{175}{5}=35\) (a, b) can be (1, 35) or (5, 7) Thus, the numbers can be (1 × 5 and 35 × 5) or (5 × 5 and 5 × 7), i.e., (5 and 175) or (25 and 35) The sum = 60 is satisfied by the pair (25, 35). Hence, the difference of the numbers is 10. |
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| 24. |
Write all the twin prime numbers from 51 to 100. |
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Answer» 1. 59 and 61 2. 71 and 73 |
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| 25. |
HCF, LCM of two numbers are 9 and 54 respectively. If one of those two numbers is 18, find the other number. |
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Answer» Given, HCF of two numbers = 9 LCM of two numbers = 54 One of the two numbers a = 18 Then, the other number b = ? We know, that the product of two numbers = LCM × HCF a × b = LCM × HCF 18 × b = 54 × 9 b = \(\frac{54\times9}{18}\) = 27 ∴ The other number = 27 |
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| 26. |
What is the LCM and HCF of twin prime numbers ? |
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Answer» LCM = Product of the taken twin primes and HCF = 1 |
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| 27. |
Explain the method of accounting for settlement of accounts at the time of dissolution. |
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Answer» Accounting Treatment in Case of Dissolution : With the dissolution of firm normal business activity stops and process of realization of firm’s assets and payment of liabilities starts. To complete this process following accounts are prepared :
Realisation account is opened on the dissolution of a firm. The object of preparing this account is to determine gain (profit) or loss on the realisation of assets and payment of liabilities. Realisation account is prepared by:
Account Entries Regarding Dissolution (ii) On Transfer of Liabilities (Except Capital Account. Current Account, Reserve, P&L Account, Reserve Fund and Partners’ Loan Account) Sundry Liabilities A/c (iii) On Sale of Assets (iv) On Taking Over Some Assets by a Partner (v) On Payment of Transferred Liabilities (vi) On Taking Responsibility of Payment of Any Liability by a Partner (vii) On Payment of Unforeseen Liability or Dissolution Expenses (viii) On Transfer of Credit Balance of (Profit) Realisation Account (ix) Transfer of Debit Balance of (Loss) Realisation Account (x) Expenses Paid by a Partner (xi) On Bearing Any Realisation Expenses by a Partner as Realisation Agent. [Note : Sometimes, liabilities are not transferred to realisation account. In such a case the liabilities are directly paid only profit loss if any arising out of it is transferred to realisation account. When such liabilities along with other liabilities which are not transferred to realisation account are paid and the following accounting entries are done in the following manner.] (xii) On Payment of Liabilities over and Above the Book Value (xiii) On Payment of Liabilities Below the Book Value Sundry (xiv) On Payment of Partner’s Loan (xv) On Distribution of Old Undistributed Profit Dissolution is as follows (xvi) On Distribution of Old Undistributed Loss (xvii) On Bringing Cash by a Partner to Make up Deficiency of Capital (xviii) On Payment of Capital Account Balance to Partners [Note: Lastly, the balance in cash/bank is only as much as it is to be paid to the partners hence, with the last entry all accounts of the firm close automatically.] Partners Capital Account: After the transfer of profit or loss on realisation, undistributed profit reserves etc. to the capital account of the partners, The balance of capital account are closed in the following manner : 1. When a partner is required to bring in cash to clear off his debit balance. The entry will be 2. When a partner is paid the credit balance of his account Cash or Bank Account:Opening balance of cash and bank and all the receipts are entered on the debit side of this account and all the payments are entered on the credit side of this account. This account must be prepared and closed last of all and the total of both the sides of this account must be equal. In this way this account also helps in the verification of the arithmetical accuracy of the account. [Note: If cash balance and bank balance both are given in the balance sheet, only one account either a cash account or a bank account is prepared. If cash account is prepared an entry is passed for withdrawing the bank balance and if a bank account is prepared, the cash balance is deposited into the bank.] Other Required Accounts : On the dissolution of firm, partner’s loan Account, partners current account, reserve and undistributed losses accounts are prepared and deficiency account in case of insolvency of all partners is also prepared. |
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| 28. |
Methods of finding HCF |
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Answer» (i) Prime Factorization : Express each of the given numbers as the product of their prime factors. The HCF of the given numbers is the product of the least powers of common factors. 24 = 23 × 3, 32 = 25 ⇒ HCF (24, 32) = 23 = 8. (ii) Continued Division Method : Divide the larger number by the smaller number. If the remainder is zero, the divisor is the HCF, otherwise divide the previous divisor by the remainder last obtained. Repeat this until the remainder becomes zero. To find the HCF of more than two numbers, first find the HCF of any two numbers and then find the HCF of the result and the third number and so on. The final HCF is the required HCF. Note : The HCF of two co-prime numbers is 1, as they have no factors in common. |
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| 29. |
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required? |
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Answer» Let the persons required to build a house in 6 days be x. Days required to build a house and number of persons are in inverse proportion. ∴ 6 × x = 8 × 3 ∴ 6 x = 24 ∴ x = 4 ∴ 4 persons are required to build the house in 6 days. |
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| 30. |
Convert the following ratios into percentages. i. 15 : 25 ii. 47 : 50iii.7 / 10 |
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Answer» i. Let 15 : 25 = x % ∴ \(\frac{15}{25}\) = \(\frac{x}{100}\) ∴ x = \(\frac{15}{25}\) x 100 = 15 x 4 = 60% ∴ 15 : 25 = 60 % ii. Let 47 : 50 = x% ∴ \(\frac{47}{50}\) = \(\frac{x}{100}\) ∴ x = \(\frac{47}{50}\) x 100 = 47 x 2 = 94% 47 : 50 = 94% iii. Let \(\frac{7}{10}\) = x % ∴ \(\frac{7}{10}\) = \(\frac{x}{100}\) ∴ x = \(\frac{7}{10}\) x 100 = 7 x 100 =70% ∴ \(\frac{7}{10}\) = 70% |
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| 31. |
Check whether the following numbers are in continued proportion.i. 2, 4, 8 ii. 1, 2, 3 iii. 9, 12, 16iv. 3, 5, 8 |
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Answer» If a, b, c are in continued proportion then b2 = ac. i. 2, 4, 8 Let, a = 2, b = 4 and c = 8 Here, b2 = 42 = 16 ac = 2 x 8 = 16 ∴ b2 = ac ∴ 2, 4,8 are in continued proportion. ii. 1, 2, 3 Let, a = 1, b = 2 and c = 3 Here, b2 = 22 = 4 ac = 1 x 3 = 3 ∴ b2 ≠ ac ∴ 1, 2,3 are not in continued proportion. iii. 9, 12, 16 Let, a = 9, b = 12 and c = 16 Here, b2 = 122 = 144 ac = 9 x 16 = 144 ∴ b2 = ac ∴ 9, 12, 16 are in continued proportion. iv. 3, 5, 8 Let, a = 3, b = 5 and c = 8 Here, b2 = 52 = 25 ac = 3 x 8 = 24 ∴ b2 ≠ ac ∴ 3, 5, 8 are not in continued proportion. |
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| 32. |
a, b, c are in continued proportion. If a = 3 and c = 27, then find b. |
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Answer» a, b, c are in continued proportion. … [Given] ∴ b2 = ac ∴ b2 = 3 x 27 …[∵ a = 3 and c = 27] ∴ b2 = 81 ∴ b = 9 …[Taking square root of both sides] |
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| 33. |
If (a + b + c)(a – b + c) = a2 + b2 + c2, show that a, b, c are in continued proportion. |
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Answer» (a + b + c)(a – b + c) = a2 + b2 + c2 … [Given] ∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2 ∴ a2 – ab + ac + ab – b2 + be + ac – be + c2 = a2 + b2 + c2 ∴ a2 + 2ac – b2 + c2 = a2 + b2 + c2 ∴ 2ac – b2 = b2 ∴ 2ac = 2b2 ∴ ac = b2 ∴ b2 = ac ∴ a, b, c are in continued proportion. |
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| 34. |
If 36, 54, x are in continued proportion, find the value of x. |
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Answer» Since 25, 35, x are in continued proportion, we have: = 36: 54 :: 54: x We know that, Product of extremes = product of means = 36 × x = 54 × 54 = 36x = 2916 = x = (2916/36) = x = 81 |
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| 35. |
If (a + b + c)(a – b + c) = a2 + b2 + c2 , show that a, b, c are in continued proportion. |
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Answer» (a + b + c)(a – b + c) = a2 + b2 + c2 …[Given] ∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2 ∴ a2 – ab + ac + ab – b2 + be + ac – be + c2 = a2 + b2 + c2 ∴ a2 + 2ac – b2 + c2 = a2 + b2 + c2 ∴ 2ac – b2 = b2 ∴ 2ac = 2b2 ∴ ac = b2 ∴ b2 = ac ∴ a, b, c are in continued proportion. |
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| 36. |
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers. |
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Answer» Let the first number be x. ∴ Third number = 26 – x 12 is the mean proportional of x and (26 – x). ∴ \(\frac{x}{12}\) = \(\frac{12}{26-x}\) ∴ x(26 – x) = 12 x 12 ∴ 26x – x2 = 144 ∴ x2 – 26x + 144 = 0 ∴ x2 – 18x – 8x + 144 = 0 ∴ x(x – 18) – 8(x – 18) = 0 ∴ (x – 18) (x – 8) = 0 ∴ x = 18 or x = 8 ∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18 ∴ The numbers are 18, 12, 8 or 8, 12, 18. |
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| 37. |
A 150m long train is moving with constant velocity of 12.5 m/s. Find (i) The equation of the motion of the train, (ii) Time taken to cross a pole, (iii) The time taken to cross the bridge of length 850 m is? |
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Answer» (i) Now m = y/x = 12.5m/second, The equation of the line is y = mx + c … (1) Put c = -150, m = 12.5 m, The equation of motion of the train is y = 12.5x – 150 (ii) To find the time taken to cross a pole we take y = 0 in (1) ⇒ 0 = 12.5x – 150 ⇒ 12.5x = 150 x = 150/12.5 = 12 seconds (iii) When y = 850 in (1) 850 = 12.5 x – 150 ⇒ 12.5x = 850 + 150 = 1000 x = 100/12.5 = 80 seconds |
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| 38. |
Prove that the straight lines joining the origin to the points of intersection of 3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angles. |
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Answer» Homogenizing the given equations 3x2 + 5xy – 3y2 + 2x + 3y = 0 and 3x – 2y – 1 = 0 (i.e) 3x – 2y = 1. We get (3x2 + 5xy – 3y2) + (2x + 3y)( 1) = 0 (i.e) (3x2 + 5xy – 3y2) + (2x + 3y)(3x – 2y) = 0 3x2 + 5xy – 3y2 + bx2 – 4xy + 9xy – 6y2 = 0 9x2 + 10xy – 9y2 = 0 Coefficient of x2 + coefficient of y2 = 9 – 9 = 0 ⇒ The pair of straight lines are at right angles. |
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| 39. |
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is …(a) 3/2(b) 9/2(c) 2/3(d) 2/9 |
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Answer» (b) 9/2 Equation of a line perpendicular to 2x – 3y + 1 = 0 is 3x + 2y = k. It passes through (1, 3). 3 + 6 = k ⇒ k = 9, 3x + 2y = 9 To find y-intercept x = 0, 2y = 9, y = 9/2 |
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| 40. |
Show that 9x2 + 24xy + 16y2 + 21x + 28y + 6 = 0 represents a pair of parallel straight lines and find the distance between them. |
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Answer» 9x2 + 24xy + 16y2 + 21x + 28y + 6 = 0 Here a = 9.6, b = 16, g = 21/2, f = 14, c = 6, h = 12 h2 – ab = (12)2 – 9(16) = 144 – 144 = 0 ∴ The lines are parallel. 9x2 + 24xy + 16y2 = (3x + 4y)(3x + 4y) Let 9x2 + 24xy + 16y2 + 21x + 28y + 6 = (3x + 4y + l)(3x + 4y + m) Equating the coefficients of x and constant term 3l + 3m = 21 lm = 6 Solving we get, l = 1 or 6 m = 6 or 1 ∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 = 0 The distance between the parallel lines are |(6 - 1)/√(9 + 16)| = 5/5 = 1 unit. |
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| 41. |
Explain the features of Micro Economics. |
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Answer» Micro economics deals with the study of economic behaviour of small individual economic units such as individual consumer, firm, producer etc. The various features of Micro economics are as follows: i. Price Theory:
ii. Partial equilibrium:
iii. Microscopic approach:
iv. Analysis of Resource Allocation and Economic Efficiency:
Who will produce the goods? What goods will be produced? In what quantities the goods will be produced? How to price the produced goods? How will they be distributed? etc.
v. Use of Marginalism Principle:
vi. Analysis of Market Structure:
vii. Based on Certain Assumptions:
viii. Limited Scope:
ix. Study of individual units:
x. Slicing Method:
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| 42. |
Micro economics is known as income theory. |
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Answer» No, I do not agree with the above statement. Reasons:
Therefore, Micro economics is not known as income theory but is known as price theory. |
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| 43. |
When did Santosh leave home for Delhi and why ? |
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Answer» When she turned sixteen and was under pressure to get married, Santosh threatened her parents that she would never marry if she did not get a proper education. Therefore, she left home and got herself enrolled in a school in Delhi. |
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| 44. |
How did Santosh begin to climb mountains ? |
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Answer» From her room in Kasturba Hostel, she watched people going up the Aravalli Hills and vanishing after a while. On investigating, she found nobody except a few mountaineers, and she asked if she could join them. They agreed and even motivated her to take to climbing and so she later accompanied them on their climbing expedition. This was how she began climbing mountains. |
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| 45. |
Why was Santosh sent to the local school ? (3) |
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Answer» Even though Santosh’s parents could afford to send their children to the best schools, she was sent to the local village school due to the’ prevailing custom in the family. |
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| 46. |
Give an example to show that even as a young girl Santosh was not ready to accept anything unreasonable. (2) |
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Answer» Santosh, from the very beginning, lived life on her own terms. She was not content with the traditional way of life and was not ready to accept anything unreasonable. Where other girls wore traditional Indian dresses, Santosh preferred shorts. |
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| 47. |
Why was the ‘holy man’ who gave Santosh’s mother his blessings surprised ? |
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Answer» The ‘holy man’ was surprised because he had assumed that Santosh’s mother wanted a son. But her grandmother told him that they did not want a son. |
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| 48. |
State the various indices to measure development of a nation. |
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Answer» The various indices to measure development of a nation are;
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| 49. |
The most important objective of the countries of the world especially after World war-II has been(A) Economic development(B) Economic gain(C) Social welfaie(D) Economic and political stability |
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Answer» Correct option is (A) Economic development |
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| 50. |
State the limitations of Human Development Index. |
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Answer» Limitations of HDI:
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