For Parents For Educators & Experts For Institutions For Employers

SSC GD Constable 13 February 2025 · Shift 1 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 108 of 210

13 February 2025 · Shift 1

Elementary Mathematics

20 questions ~20 min readFree

20 questions

400 apples were bought at ₹1210 per hundred and were sold at a profit of ₹860. Find the selling price (in ₹) per dozen of apples.

A.181
B.161
C.171
D.186

Solution

Profit and selling price The rate is ₹1210 per hundred apples, so 400 apples cost

\frac{400}{100} \times 1210 = 4 \times 1210

, i.e. ₹4,840. Selling price = cost price + profit =4840+860=5700, i.e. ₹5,700. 400 apples

= \frac{400}{12} = \frac{100}{3}

dozen. Selling price per dozen

= 5700 \div \frac{100}{3} = 5700 \times \frac{3}{100} = 171

. Hence the selling price is ₹171 per dozen, so option (c) is correct.

Ashish has two grandsons Ketan and Mahesh. 15 year old Ketan gets some money from Ashish's wealth and 16 year old Mahesh gets the rest of the money. But Ketan and Mahesh will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate 4% compounded annually. When both turn 23 , they receive the same amount. How much had Ashish given Mahesh (in ₹) initially, if total money with Ashish was ₹25500?

A.12500
B.12250
C.13350
D.13000

Solution

Compound interest division Ketan is 15, so his share is invested for 23-15=8 years; Mahesh is 16, so his share is invested for 23-16=7 years. Let Ketan get ₹K and Mahesh get ₹M. Equal maturity amounts give

K {(1.04)}^{8} = M {(1.04)}^{7}

Cancelling

{(1.04)}^{7}

from both sides:

M = 1.04 K = \frac{26}{25} K

K + M = 25500 \Rightarrow K (1 + \frac{26}{25}) = 25500 \Rightarrow K \times \frac{51}{25} = 25500

K = \frac{25500 \times 25}{51} = 12500

, so M=25500-12500=13000. Mahesh was given ₹13,000 initially, so option (d) is correct.

The price of fuel decreases by 25%, 35% and 40% in three successive months, but increases by 60% in the fourth month. What is the percentage increase/decrease in the price of fuel in the fourth month as compared to its original price?

A.Decreases by 57.28%
B.Decreases by 53.2%
C.Increases by 55.35%
D.Increases by 50.32%

Solution

Successive percentage change Successive percentage changes multiply, so take the original price as 100 and multiply the four factors.

Final price

= 100 \times \frac{75}{100} \times \frac{65}{100} \times \frac{60}{100} \times \frac{160}{100}

= 100 \times 0.75 \times 0.65 \times 0.60 \times 1.60 = 46.8

. Net change =46.8-100=-53.2 on a base of 100, i.e. a fall of 53.2%. So the price decreases by 53.2%, and option (b) is correct.

The average score of 5 players of a cricket team is 55. It is also observed that 3 of these 5 players have an average score of 55 . What will be the average score of the remaining two players ?

A.55
B.50
C.45
D.60

Solution

Average of a part group Total score of all 5 players

= 5 \times 55 = 275

. Total score of the 3 players

= 3 \times 55 = 165

. Total score of the remaining two =275-165=110. Their average

= \frac{110}{2} = 55

. Hence the required average is 55, so option (a) is correct.

The greatest number, which divides 2248 and 1761 to leave 3 and 1 respectively as remainders, is:

A.14
B.5
C.19
D.24

Solution

HCF with remainders A number that leaves remainder 3 in 2248 divides 2248-3=2245 exactly. The same number leaves remainder 1 in 1761, so it divides 1761-1=1760 exactly. The greatest such number is therefore the HCF of 2245 and 1760.

2245 = 5 \times 449

and

1760 = 2^{5} \times 5 \times 11

, and 449 is prime. The only common factor is 5, so the HCF =5 (and 3<5, 1<5, so both remainders are legal). Hence the greatest number is 5, so option (b) is correct.

The speed of a train is 65 km per hour. In 35 minutes and 15 seconds, the train will cover ____ km and ____ metres.

A.36; 187.5
B.38; 187.5
C.36; 178.5
D.38; 178.5

Solution

Speed, time and distance First put the time in hours: 35 minutes 15 seconds

= 35 + \frac{15}{60} = 35.25

minutes

= \frac{35.25}{60}

hours. Distance = speed × time

= 65 \times \frac{35.25}{60}

= \frac{2291.25}{60} = 38.1875

km. The whole-number part is 38 km, and 0.1875 km

= 0.1875 \times 1000 = 187.5

m. So the train covers 38 km and 187.5 metres, and option (b) is correct.

Two trains are moving in opposite directions at speeds of

50 km / h

and 80

km / h

. The length of one train is 480 m . The time taken by them to cross each other is 28 seconds. The length (in m ) of the other train, correct to 2 decimal places, is:

A.533.51
B.530.12
C.529.39
D.531.11

Solution

Trains crossing each other Moving in opposite directions, the speeds add: relative speed =50+80=130 km/h. 130 km/h

= 130 \times \frac{5}{18} = \frac{325}{9}

m/s. While crossing each other the trains together cover the sum of their two lengths. Sum of lengths

= \frac{325}{9} \times 28 = \frac{9100}{9} = 1011.11

m. Length of the other train =1011.11-480=531.11 m. So option (d) is correct.

Ramu invested some money in HDFC at 4% per annum rate of interest. What would be the corresponding simple interest (in ₹) if after 2 years, Ramu got ₹127.5 as compound interest, considering annual compounding ?

A.135
B.120
C.125
D.140

Solution

Compound and simple interest For 2 years at 4% per annum, CI on a principal P is

P ({(1 + \frac{4}{100})}^{2} - 1)

{(1.04)}^{2} - 1 = 1.0816 - 1 = 0.0816

, i.e. an effective 8.16% over the two years. So

0.0816 P = 127.5 \Rightarrow P = \frac{127.5}{0.0816} = 1562.5

. Simple interest

= \frac{P \times R \times T}{100} = \frac{1562.5 \times 4 \times 2}{100} = 125

. Hence the corresponding simple interest is ₹125, so option (c) is correct.

A and B can do a piece of work in 28 days;

B

and

C

can do the same work in 18 days and

A, B

and C together can do it in 16 days. The time required (in days) to finish the work by A and C is:

A. $29 \frac{11}{17}$
B. $25 \frac{11}{13}$
C. $23 \frac{11}{15}$
D. $27 \frac{13}{19}$

Solution

Time and work, pairs Take the total work as the LCM of 28, 18 and 16, which is 1008 units.

Then

A + B = \frac{1008}{28} = 36

B + C = \frac{1008}{18} = 56

and

A + B + C = \frac{1008}{16} = 63

units per day. So C=63-36=27 units per day and A=63-56=7 units per day. A+C=7+27=34 units per day. Time

= \frac{1008}{34} = \frac{504}{17} = 29 \frac{11}{17}

days. Hence option (a) is correct.

The third proportion of numbers a and b is ab . Which of the following statements is true?

A. $2 a = b$
B. $a = 2 b$
C. $a^{2} = b$
D. $a = b^{2}$

Solution

Third proportional If x is the third proportional to a and b, then a:b=b:x, i.e.

a x = b^{2}

. So

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

. The question says this third proportional is ab, so

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

. Cross-multiplying,

b^{2} = a^{2} b

, and dividing by b (as

b \neq 0

) gives

b = a^{2}

. Hence

a^{2} = b

, so option (c) is correct.

When

12, 15, 18, 20

and 25 divide the least number

x

, the remainder in each case was 3 but

x

was divisible by 7 . What is the value of

x

A.390
B.930
C.903
D.309

Solution

LCM with common remainder A number leaving remainder 3 with each of 12, 15, 18, 20 and 25 is of the form (a multiple of their LCM)

+ 3

12 = 2^{2} \times 3

15 = 3 \times 5

18 = 2 \times 3^{2}

20 = 2^{2} \times 5

25 = 5^{2}

, so the LCM

= 2^{2} \times 3^{2} \times 5^{2} = 900

. So x=900k+3 for

k = 1, 2, 3, \dots

Taking k=1 gives x=903, and

903 = 7 \times 129

, so it is divisible by 7 as required. Since k=1 is the smallest value, the least such x is 903, so option (c) is correct.

Which of the following schemes provide the highest discount?

A.Two successive discounts of 10% and 3%
B.Two successive discounts of 10% each
C.Two successive discounts of 7% and 10%
D.Two successive discounts of 5% and 10%

Solution

Successive discounts Two successive discounts leave

\frac{100 - x}{100} \times \frac{100 - y}{100}

of the marked price, where x and y are the two discount rates.

(a)

0.90 \times 0.97 = 0.873

, a net discount of 12.7%. (b)

0.90 \times 0.90 = 0.81

, a net discount of 19%. (c)

0.93 \times 0.90 = 0.837

, a net discount of 16.3%. (d)

0.95 \times 0.90 = 0.855

, a net discount of 14.5%. The scheme that leaves the least price gives the greatest discount, so two discounts of 10% each are best and option (b) is correct.

Rajesh bought 5 kg of apples for ₹875 and sold it for ₹220 per kg. How much is the profit gained by him?

A.₹220
B.₹215
C.₹225
D.₹200

Solution

Profit on a purchase Cost price of 5 kg of apples = ₹875. Selling price

= 5 \times 220 = 1100

, i.e. ₹1,100. Profit = selling price − cost price =1100-875=225. Hence the profit is ₹225, so option (c) is correct.

The fourth proportional of a, c and bc is d. The fourth proportional of bc²/a, ad/c and c is ____ .

Solution

Fourth proportional If d is the fourth proportional to a, c and bc, then a:c=bc:d, so

a d = c \times b c

and

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

. Let x be the fourth proportional to

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

\frac{a d}{c}

and c, so

\frac{b c^{2}}{a} : \frac{a d}{c} = c : x

. Then

\frac{b c^{2}}{a} \times x = \frac{a d}{c} \times c = a d

, so

x = \frac{a d \times a}{b c^{2}} = \frac{a^{2} d}{b c^{2}}

. Substituting

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

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

. Hence the blank is a, so option (b) is correct.

Find the value of [(22 × 6) × {4 ÷ 4 × (19 − 12) ÷ 7}]

A.116
B.150
C.132
D.117

Solution

BODMAS simplification By BODMAS, clear the innermost bracket first: 19-12=7. Inside the curly bracket, division and multiplication are done left to right:

4 \div 4 \times 7 \div 7 = 1 \times 7 \div 7 = 1

. Inside the round bracket:

22 \times 6 = 132

. So the value is

132 \times 1 = 132

, and option (c) is correct.

The ratio of the areas of two triangles is 4 : 3 and the ratio of their heights is

3 : 4

. The ratio of their bases is:

A.16 : 9
B.3 : 2
C.4 : 9
D.1 : 6

Solution

Ratio of triangle bases Area of a triangle

= \frac{1}{2} \times b \times h

, so

b = \frac{2 A}{h}

- the base grows with the area and shrinks as the height grows.

Therefore

\frac{b_{1}}{b_{2}} = \frac{A_{1}}{A_{2}} \times \frac{h_{2}}{h_{1}} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}

. So the bases are in the ratio 16 : 9, and option (a) is correct.

The LCM of two numbers is 126 times their HCF. If their HCF is 7 and the difference between the two numbers is five times the HCF, then the sum of the two numbers is:

A.611
B.218
C.161
D.116

Solution

HCF, LCM and the numbers Since the HCF is 7, write the numbers as 7a and 7b, where a and b are co-prime. LCM

= 126 \times

HCF

= 126 \times 7 = 882

. Product of the numbers = HCF × LCM

= 7 \times 882 = 6174

, so 49ab=6174 and ab=126. Difference

= 5 \times 7 = 35

, so 7(a-b)=35 and a-b=5.

Among the factor pairs of 126, only

a = 14, b = 9

gives a-b=5, and 14 and 9 are indeed co-prime. The numbers are

7 \times 14 = 98

and

7 \times 9 = 63

, so their sum =98+63=161 and option (c) is correct.

The average weight of Gopal, Akshay and Atul is 45 kg . If the average weight of Gopal and Akshay be 43 kg and that of Akshay and Atul be 43 kg, then the weight of Akshay (in kg) is:

A.37
B.57
C.47
D.52

Solution

Averages of overlapping pairs Gopal + Akshay + Atul

= 3 \times 45 = 135

kg. Gopal + Akshay

= 2 \times 43 = 86

kg, so Atul =135-86=49 kg. Akshay + Atul

= 2 \times 43 = 86

kg, so Akshay =86-49=37 kg. Hence Akshay weighs 37 kg, so option (a) is correct.

Shivam invests a sum of ₹5400 and Rupesh invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 4 years, Rupesh gets ₹600 more interest than Shivam, then find the rate of interest per annum (in percentage).

A.2.75
B.5.75
C.3.75
D.1.75

Solution

Simple interest, rate At the same rate for the same time, simple interest is proportional to the principal, so the extra ₹600 comes only from the extra principal. Extra principal =9400-5400=4000.

\frac{4000 \times R \times 4}{100} = 600

160 R = 600 \Rightarrow R = \frac{600}{160} = 3.75

. Hence the rate is 3.75% per annum, so option (c) is correct.

There are 1224 employees in an organisation, out of which 25% are promoted. How many such employees are there who got promotion ?

A.312
B.306
C.308
D.310

Solution

Percentage of a total 25% is the same as

\frac{1}{4}

, so promoted employees

= \frac{25}{100} \times 1224 = \frac{1224}{4}

\frac{1224}{4} = 306

. Hence 306 employees got promotion, so option (b) is correct.

Finished reading? Now test yourself.

The same 20 questions, but with a timer and the answers hidden — 15 minutes.

Take the test

Full contents list