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SSC GD Constable 11 February 2025 · Shift 3 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 88 of 210

11 February 2025 · Shift 3

Elementary Mathematics

20 questions ~20 min readFree

20 questions

A shopkeeper bought an article for ₹520. At what price (in ₹) should he sell the article to make 30% profit?

A.364
B.688
C.676
D.664

Solution

Profit and selling price Cost price (CP) = ₹520 and the required profit = 30%. Formula:

SP = CP \times \frac{100 + p}{100}

, where p is the profit percent.

SP = 520 \times \frac{100 + 30}{100} = 520 \times \frac{13}{10}

SP = 52 \times 13 = 676

So the article must be sold for ₹676. Hence option (c) is correct.

Find the value of [(12 × 10) × {6 ÷ 6 × (15 − 12)/3}]

A.120
B.138
C.122
D.109

Solution

BODMAS simplification Simplify from the innermost bracket outwards, following BODMAS. Innermost bracket:

\frac{(15 - 12)}{3} = \frac{3}{3} = 1

Inside the braces:

6 \div 6 \times 1 = 1 \times 1 = 1

Outside them:

12 \times 10 = 120

(120 \times 1) = 120

So the value is 120. Hence option (a) is correct.

The price (per litre) of petrol increases by 75%. By what percent should its consumption be reduced such that the expenditure on it increases by 47% only?

A.16%
B.77%
C.26%
D.84%

Solution

Percentage - price and consumption Expenditure = Price × Consumption, so start with price = 100 and consumption = 100 units. Original expenditure

= 100 \times 100 = 10000

After the rise of 75% the new price =100+75=175 Expenditure is allowed to rise by 47% only, so it becomes

10000 \times \frac{147}{100} = 14700

New consumption

= \frac{14700}{175} = 84

units Fall in consumption =100-84=16 units, i.e.

\frac{16}{100} \times 100 = 16

So the consumption must be reduced by 16%. Hence option (a) is correct.

The marked price of an electronic watch in a store is ₹45,880 and it is available at a discount of 25%. What is the price (in ₹) that a customer pays if he buys it from the store?

A.31,440
B.34,410
C.31,400
D.34,140

Solution

Discount and selling price Marked price (MP) = ₹45,880 and the discount = 25%. Formula:

SP = MP \times \frac{100 - d}{100}

, where d is the discount percent.

SP = 45880 \times \frac{100 - 25}{100} = 45880 \times \frac{3}{4}

SP = 11470 \times 3 = 34410

So the customer pays ₹34,410. Hence option (b) is correct.

The perimeter of a rectangle is 26 cm and its length is 7 cm. Find its area in cm².

A.32
B.42
C.36
D.48

Solution

Perimeter and area of a rectangle Formula: perimeter of a rectangle =2(l+b), where l is the length and b the breadth. 2(7+b)=26

7 + b = 13 \Rightarrow b = 6

Area

= l \times b = 7 \times 6 = 42

So the area is

42 {cm}^{2}

. Hence option (b) is correct.

The greatest number that will divide 42, 84 and 153 so as to leave the same remainder in each case is:

Solution

HCF - equal remainders A number that leaves the same remainder on dividing several numbers must divide their differences exactly. 84-42=42, 153-84=69, 153-42=111 Required number

= HCF (42, 69, 111)

42 = 2 \times 3 \times 7

69 = 3 \times 23

111 = 3 \times 37

The only factor common to all three is 3, so the HCF is 3. So the greatest such number is 3. Hence option (c) is correct.

Anil has ₹1224 with him. He divided it amongst his sons Ashok and Raj and asked them to invest it at 4% rate of interest compounded annually. It was seen that Ashok and Raj got same amount after 13 and 14 years respectively. How much (in ₹) did Anil give to Raj ?

A.450
B.624
C.600
D.724

Solution

Compound interest - equal amounts Both sons end with the same amount, and Raj's money grows for one year longer, so Raj must be given less. Formula: Amount

= P {(1 + \frac{R}{100})}^{T}

P_{A} {(1 + \frac{4}{100})}^{13} = P_{R} {(1 + \frac{4}{100})}^{14}

\frac{P_{A}}{P_{R}} = 1 + \frac{4}{100} = \frac{26}{25}

So Ashok : Raj =26:25 and the total is 26+25=51 units. 51 units = ₹1224, so 1 unit = ₹24.

Raj's share

= 25 \times 24 = 600

So Anil gave ₹600 to Raj. Hence option (c) is correct.

A can do a piece of work in 15 days and B can do it in 24 days. If they do the work alternately, in how many days will they finish the work, provided that A begins the work?

A. $18 \frac{1}{4}$
B. $18 \frac{3}{8}$
C.17
D. $17 \frac{2}{5}$

Solution

Alternate-day work Take the total work as

LCM (15, 24) = 120

units. A's one-day work

= \frac{120}{15} = 8

units and B's one-day work

= \frac{120}{24} = 5

units. A begins, so each pair of days (A then B) finishes 8+5=13 units.

9 pairs take 18 days and complete

9 \times 13 = 117

units. Work left =120-117=3 units, and the next turn is A's. Time A needs for it

= \frac{3}{8}

day. Total time

= 18 + \frac{3}{8} = 18 \frac{3}{8}

days. Hence option (b) is correct.

The simple interest on a principal amount (in ₹) is ₹284 for a period of 2 years at the rate of 5% per annum. The principal amount (in ₹) is:

A.2832
B.2840
C.2844
D.2845

Solution

Simple interest - principal Formula:

SI = \frac{P \times R \times T}{100}

284 = \frac{P \times 5 \times 2}{100}

284 = \frac{10 P}{100} = \frac{P}{10}

P = 284 \times 10 = 2840

So the principal is ₹2840. Hence option (b) is correct.

The average weight of Atul, Baban and Rakesh is 46 kg . If the average weight of Atul and Baban be 39 kg and that of Baban and Rakesh be 50 kg , then the weight of Baban (in kg) is:

A.60
B.50
C.55
D.40

Solution

Averages - weights Total = average × number of persons. Atul + Baban + Rakesh

= 46 \times 3 = 138

Atul + Baban

= 39 \times 2 = 78

Baban + Rakesh

= 50 \times 2 = 100

Adding the last two: Atul

+ 2 \times

Baban + Rakesh =78+100=178 Subtracting the total of all three: Baban =178-138=40 So Baban weighs 40 kg. Hence option (d) is correct.

On a circular path of 432 m, Soham and Rohan start running from the same point and in same direction at the speed of 12 m/sec and 9 m/sec, respectively. Find the time in which they will meet for the first time.

A.98 seconds
B.144 seconds
C.156 seconds
D.112 seconds

Solution

Circular track - first meeting Running in the same direction, the faster runner meets the slower one only after gaining one full lap on him. Relative speed =12-9=3 m/sec Formula: time

= \frac{length of the track}{relative speed}

Time

= \frac{432}{3} = 144

So they meet for the first time after 144 seconds. Hence option (b) is correct.

Find the value of [(14 × 9) × {8 ÷ 2 × (18 − 15)/3}]

A.517
B.504
C.521
D.513

Solution

BODMAS simplification Simplify from the innermost bracket outwards, following BODMAS. Innermost bracket:

\frac{(18 - 15)}{3} = \frac{3}{3} = 1

Inside the braces:

8 \div 2 \times 1 = 4 \times 1 = 4

Outside them:

14 \times 9 = 126

(126 \times 4) = 504

So the value is 504. Hence option (b) is correct.

Akshay has ₹1218 with him. He divided it amongst his sons Arjun and Kartik and asked them to invest it at 10% rate of interest compounded annually. It was seen that Arjun and Kartik got same amount after 12 and 13 years respectively. How much (in ₹) did Akshay give to Kartik?

A.430
B.738
C.580
D.638

Solution

Compound interest - equal amounts Both amounts are equal, and Kartik's money grows for one year longer, so Kartik gets the smaller share. Formula: Amount

= P {(1 + \frac{R}{100})}^{T}

P_{A} {(1 + \frac{10}{100})}^{12} = P_{K} {(1 + \frac{10}{100})}^{13}

\frac{P_{A}}{P_{K}} = 1 + \frac{10}{100} = \frac{11}{10}

So Arjun : Kartik =11:10 and the total is 11+10=21 units. 21 units = ₹1218, so 1 unit = ₹58.

Kartik's share

= 10 \times 58 = 580

So Akshay gave ₹580 to Kartik. Hence option (c) is correct.

A number is increased by 50% and then again by 50%. By what percentage should the increased number be reduced so as to get back the original number ?

A.40 2/3%
B.55 5/9%
C.50 1/5%
D.45 4/3%

Solution

Successive percentage change Let the original number be 100. After the first increase of 50%:

100 \times \frac{3}{2} = 150

After the second increase of 50%:

150 \times \frac{3}{2} = 225

To get 100 back, the increased number must fall by 225-100=125. Required reduction

= \frac{125}{225} \times 100 = \frac{500}{9}

\frac{500}{9} = 55 \frac{5}{9}

So the increased number must be reduced by

55 \frac{5}{9}

percent. Hence option (b) is correct.

Find the mean proportional between 6 and 150 .

A.90
B.45
C.60
D.30

Solution

Mean proportional The mean proportional between a and b is

\sqrt{a b}

, because a:x::x:b gives

x^{2} = a b

x = \sqrt{6 \times 150} = \sqrt{900}

x=30 So the mean proportional is 30. Hence option (d) is correct.

400 mangoes were bought at ₹1210 per hundred and were sold at a profit of ₹960. Find the selling price (in ₹) per dozen of mangoes.

A.164
B.174
C.184
D.189

Solution

Profit - rate per dozen 100 mangoes cost ₹1210, and 400 mangoes are four such hundreds. Cost price of 400 mangoes

= 4 \times 1210 = 4840

Selling price of 400 mangoes =4840+960=5800 Selling price of one mango

= \frac{5800}{400} = 14.5

Selling price per dozen

= 14.5 \times 12 = 174

So one dozen sells for ₹174. Hence option (b) is correct.

Find the ratio between the third proportional of 12 and 42 with the third proportional of 16 and 56.

A.3 : 4
B.2 : 3
C.5 : 6
D.4 : 5

Solution

Third proportional If c is the third proportional to a and b, then a:b::b:c, so

c = \frac{b^{2}}{a}

. For 12 and 42:

c_{1} = \frac{42^{2}}{12} = \frac{1764}{12} = 147

For 16 and 56:

c_{2} = \frac{56^{2}}{16} = \frac{3136}{16} = 196

Required ratio =147:196 Dividing both by 49: 147:196=3:4 So the ratio is 3 : 4. Hence option (a) is correct.

The time period of a simple pendulum is found to be 2.2 sec, 1.8 sec, 2 sec, 1.6 sec and 2.4 sec. What is the average time period (in sec)?

A.2.4
B.2.2
C.2
D.1.6

Solution

Average of readings Formula: average

= \frac{sum of the readings}{number of readings}

Sum =2.2+1.8+2+1.6+2.4=10 There are 5 readings, so average

= \frac{10}{5} = 2

So the average time period is 2 sec. Hence option (c) is correct.

Two trains having lengths of 300 m and 200 m are running at speeds of 70 km/h and 120 km/h, respectively, in the same direction. The time taken (in minutes) by the faster train, coming from behind, to completely cross the other train is:

A.3
B.2
C.3.9
D.0.6

Solution

Trains crossing - same direction To cross completely, the faster train must cover the sum of the two lengths. Distance =300+200=500 m Both move the same way, so relative speed =120-70=50 km/h.

50 km / h = 50 \times \frac{5}{18} = \frac{125}{9}

m/sec Time

= \frac{500}{\frac{125}{9}} = 500 \times \frac{9}{125} = 36

seconds 36 seconds

= \frac{36}{60} = 0.6

minute So the faster train takes 0.6 minute. Hence option (d) is correct.

When 7516, 7635 and 7992 are divided by the greatest number x, then the remainder in each case is the same. The product of the digits of x is:

Solution

HCF - equal remainders The remainder is the same each time, so x divides the differences of the numbers exactly. 7635-7516=119, 7992-7635=357, 7992-7516=476

x = HCF (119, 357, 476)

119 = 7 \times 17

357 = 3 \times 7 \times 17

476 = 4 \times 7 \times 17

HCF = 7 \times 17 = 119

, so x=119. Product of the digits of 119

= 1 \times 1 \times 9 = 9

So the required product is 9. Hence option (a) is correct.

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