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SSC GD Constable 20 February 2025 · Shift 2 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 173 of 210

20 February 2025 · Shift 2

Elementary Mathematics

20 questions ~20 min readFree

20 questions

Evaluate:

(- 9) - (- 60) \div (- 12) + (- 3) \times 6

A.-35
B.-34
C.-32
D.-31

Solution

BODMAS simplification BODMAS order: do the division first, then the multiplication, and only then the addition and subtraction. Division:

(- 60) \div (- 12) = + 5

, because a negative divided by a negative is positive. Multiplication:

(- 3) \times 6 = - 18

. The expression becomes (-9)-(+5)+(-18). =-9-5-18=-32 Hence the correct option is (c).

The LCM of

46, 68, 136

and 138 is:

A.9329
B.9455
C.9384
D.9406

Solution

LCM by prime factors Break every number into prime factors first.

46 = 2 \times 23

68 = 2^{2} \times 17

136 = 2^{3} \times 17

138 = 2 \times 3 \times 23

The LCM keeps the highest power of each prime that occurs:

2^{3} \times 3 \times 17 \times 23

= 8 \times 3 \times 17 \times 23 = 24 \times 391 = 9384

Hence the correct option is (c).

If the cost price of 28 cricket bats is the same as the selling price of 32 cricket bats, then the loss percentage is:

A.14.3%
B.32%
C.12.5%
D.28%

Solution

Loss percentage Let the cost price of one bat be ₹1, so the CP of 28 bats =₹28. This equals the SP of 32 bats, so the SP of one bat

= \frac{28}{32} = ₹ \frac{7}{8}

. SP is less than CP, so there is a loss of

1 - \frac{7}{8} = ₹ \frac{1}{8}

on each bat. Loss

% = \frac{Loss}{CP} \times 100 = \frac{1 / 8}{1} \times 100 = 12.5 %

Shortcut: when CP of a articles = SP of b articles with b>a, loss

% = \frac{b - a}{b} \times 100 = \frac{32 - 28}{32} \times 100 = 12.5 %

. Hence the correct option is (c).

There are 44 members in group

A

, 26 members in group B and 45 members in group C . All the members of these groups went to a restaurant. The average amount spent on each member of group A, B and C is ₹403, ₹383 and ₹228, respectively. The total average amount (in ₹) spent per member is:

A.330
B.334
C.329
D.327

Solution

Combined average Formula: combined average

= \frac{total amount}{total members}

, so first turn each average back into a total. Group A:

44 \times 403 = 17732

; Group B:

26 \times 383 = 9958

; Group C:

45 \times 228 = 10260

Total amount =17732+9958+10260=₹37950 Total members =44+26+45=115 Required average

= \frac{37950}{115} = ₹ 330

Hence the correct option is (a).

Atul invests a sum of ₹5400 and Nirmal invests a sum of ₹10200 at the same rate of simple interest per annum. If, at the end of 5 years, Nirmal gets ₹480 more interest than Atul, then find the rate of interest per annum (in percentage).

Solution

Simple interest rate The rate and the time are the same for both, so the extra interest comes only from the extra principal. Extra principal =10200-5400=₹4800 Formula:

S I = \frac{P \times R \times T}{100}

\frac{4800 \times R \times 5}{100} = 480

240R=480

R = \frac{480}{240} = 2 %

per annum Hence the correct option is (a).

A cylindrical tank has a capacity of

4774 m^{3}

. If the radius of the base is 7 m , what is the depth? (Use

π = \frac{22}{7}

)

A.32 m
B.33 m
C.30 m
D.31 m

Solution

Volume of a cylinder The capacity of the tank is the volume of a cylinder, and its depth is the height h.

Formula:

V = π r^{2} h

\frac{22}{7} \times 7^{2} \times h = 4774

\frac{22}{7} \times 49 = 154

, so 154h=4774

h = \frac{4774}{154} = 31

m Hence the correct option is (d).

400 apples were bought at

₹ 1220

per hundred and were sold at a profit of ₹920. Find the selling price (in ₹) per dozen of apples.

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

Solution

Profit and selling price 400 apples are

\frac{400}{100} = 4

hundreds, so the cost price

= 4 \times 1220 = ₹ 4880

. Selling price = cost price + profit =4880+920=₹5800 for all 400 apples. Selling price of one apple

= \frac{5800}{400} = ₹ 14.50

A dozen is 12 apples, so SP per dozen

= 14.50 \times 12 = ₹ 174

. Hence the correct option is (c).

The price of fuel decreases by 15%, 25% and 35% in three successive months, but increases by 40% 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 41.98%
B.Decreases by 40.3%
C.Increases by 39.36%
D.Increases by 45.33%

Solution

Successive percentage change Successive changes multiply, they do not add, so take the original price as ₹100 and apply one factor per month. A fall of

15 %

25 %

and

35 %

multiplies by 0.85, 0.75 and 0.65; a rise of

40 %

multiplies by 1.40.

Final price

= 100 \times 0.85 \times 0.75 \times 0.65 \times 1.40 = ₹ 58.0125

Since 58.0125<100, the price has fallen. Fall =100-58.0125=₹41.9875 Decrease

% = \frac{41.9875}{100} \times 100 = 41.9875 % \approx 41.98 %

Hence the correct option is (a).

If 4% of x=96, then x is equal to:

A.4800
B.2400
C.4900
D.2500

Solution

Percentage of a number "

4 %

of x" means

\frac{4}{100} \times x

\frac{4 x}{100} = 96

4 x = 96 \times 100 = 9600

x = \frac{9600}{4} = 2400

Hence the correct option is (b).

LCM and HCF of two numbers are 198 and 33 respectively. If one of the numbers is 99 , then find the other number.

A.63
B.67
C.68
D.66

Solution

LCM and HCF relation Formula: LCM

\times

HCF = product of the two numbers.

198 \times 33 = 99 \times other number

Other number

= \frac{198 \times 33}{99} = 2 \times 33 = 66

Check: for 99 and 66 the HCF is 33 and the LCM is 198, so the pair fits. Hence the correct option is (d).

Ashok has ₹1617 with him. He divided it amongst his sons Raj and Varun and asked them to invest it at 10% rate of interest compounded annually. It was seen that Raj and Varun got same amount after 18 and 19 years respectively. How much (in ₹) did Ashok give to Raj ?

A.847
B.870
C.770
D.697

Solution

Compound interest ratio Let Raj get ₹R and Varun get ₹V; Raj's money grows for 18 years and Varun's for 19 years. Formula: amount

= P {(1 + \frac{r}{100})}^{n}

, and here the two amounts are equal.

R {(1 + \frac{10}{100})}^{18} = V {(1 + \frac{10}{100})}^{19}

\frac{R}{V} = {(\frac{11}{10})}^{19 - 18} = \frac{11}{10}

So Raj : Varun =11:10, a total of 11+10=21 equal units. 21 units

= ₹ 1617 \Rightarrow 1

unit

= \frac{1617}{21} = ₹ 77

Raj's share

= 11 \times 77 = ₹ 847

Hence the correct option is (a).

In a medical store, the marked price of an injection is ₹6,490. The medical store owner gives a discount of 18% on its sale. What is the sale price (in

₹

) of the injection? (Rounded off to the nearest rupee.)

A.5,112
B.5,421
C.5,322
D.5,233

Solution

Discount and sale price A discount of

18 %

leaves

100 - 18 = 82 %

of the marked price to be paid. Formula: SP = MP

\times (1 - \frac{d}{100})

= 6490 \times \frac{82}{100}

= \frac{532180}{100} = ₹ 5321.80

Rounded to the nearest rupee, SP =₹5322. Hence the correct option is (c).

Find the value of

((17 \times 16) \times (8 \div 8 \times \frac{(18 - 11)}{7}))

A.272
B.263
C.255
D.283

Solution

Bracket simplification Work from the innermost bracket outwards. Innermost: 18-11=7, so

\frac{18 - 11}{7} = \frac{7}{7} = 1

. Inside the braces:

8 \div 8 \times 1 = 1 \times 1 = 1

. First bracket:

17 \times 16 = 272

. Value

= 272 \times 1 = 272

Hence the correct option is (a).

The fourth proportional to 17, 87 and 119 is:

A.354
B.405
C.525
D.609

Solution

Fourth proportional The fourth proportional x to a, b, c satisfies a:b::c:x, i.e.

\frac{a}{b} = \frac{c}{x}

. Cross-multiplying gives

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

x = \frac{87 \times 119}{17}

\frac{119}{17} = 7

, so

x = 87 \times 7 = 609

. Hence the correct option is (d).

Pramod has two grandsons Ashish and Ketan. 13 year old Ashish gets some money from Pramod's wealth and 14 year old Ketan gets rest of the money. But Ashish and Ketan will get money only when they turn 24 years old. Till then the money is in a bank getting interest at rate

4 %

compounded annually. When both turn 24 , they receive the same amount. How much had Pramod given Ketan (in ₹) initially, if total money with Pramod was ₹25500 ?

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

Solution

Compound interest division Ashish is 13, so his money is invested for 24-13=11 years; Ketan is 14, so his is invested for 24-14=10 years. Let Ashish get ₹A and Ketan get ₹K; at 24 the two amounts are equal.

A {(1 + \frac{4}{100})}^{11} = K {(1 + \frac{4}{100})}^{10}

\frac{A}{K} = {(\frac{26}{25})}^{10 - 11} = \frac{25}{26}

So Ashish : Ketan =25:26, a total of 25+26=51 equal units. 51 units

= ₹ 25500 \Rightarrow 1

unit

= \frac{25500}{51} = ₹ 500

Ketan's share

= 26 \times 500 = ₹ 13000

Hence the correct option is (b).

Two trains are moving in opposite directions at speeds of

60 km / h

and 70

km / h

. The length of one train is 170 m . The time taken by them to cross each other is 26 seconds. The length (in m ) of the other train, correct to 2 decimal places, is:

A.768.88
B.770.75
C.771.56
D.769.34

Solution

Trains crossing each other Let the other train be x m long. The trains move in opposite directions, so the speeds add: relative speed =60+70=130 km/h. 130 km/h

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

m/s While crossing, the distance covered is the sum of the two lengths: 170+x.

170 + x = \frac{325}{9} \times 26 = \frac{8450}{9} = 938.88

x=938.88-170=768.88 m Hence the correct option is (a).

A and B can finish a work in 20 days while

B

and

C

can do it in 36 days. A started the work, worked for 5 days, then B worked for 10 days and the remaining work was finished by C in 15 days. In how many days could C alone have finished the whole work?

A. $16 \frac{4}{11}$
B. $15 \frac{8}{11}$
C. $16 \frac{8}{11}$
D. $15 \frac{4}{11}$

Solution

Time and work Let A, B and C do a, b and c part of the work in one day, so

a + b = \frac{1}{20}

and

b + c = \frac{1}{36}

. A worked 5 days, B worked 10 days and C worked 15 days, and together they finished the job: 5a+10b+15c=1. Only the two pairs are known, so regroup the left side to use them: 5(a+b)+5(b+c)+10c=5a+10b+15c.

\frac{5}{20} + \frac{5}{36} + 10 c = 1

10 c = 1 - \frac{1}{4} - \frac{5}{36} = \frac{36 - 9 - 5}{36} = \frac{22}{36} = \frac{11}{18}

c = \frac{11}{180}

, so C alone needs

\frac{180}{11} = 16 \frac{4}{11}

days. Hence the correct option is (a).

A man travels at

50 km / hr

for 2 hours, the next at

40 km / hr

for 2.5 hours and the next at

60 km / hr

for 3 hours. What is his average speed (in km/hr, rounded off to two decimal places) for the whole journey?

A.50.67
B.48.33
C.46.67
D.53.33

Solution

Average speed Formula: average speed

= \frac{total distance}{total time}

; the three speeds must never simply be averaged. Distances:

50 \times 2 = 100

km,

40 \times 2.5 = 100

km,

60 \times 3 = 180

km Total distance =100+100+180=380 km Total time =2+2.5+3=7.5 hours Average speed

= \frac{380}{7.5} = 50.666 \dots \approx 50.67

km/hr Hence the correct option is (a).

A train travels the first 200 km of a journey at a speed of

100 km / h

and the remaining 150 km at a speed of

75 km / h

. What is the average speed (in km/h) of the train for the entire journey?

A.92.5
B.87.5
C.85
D.90

Solution

Average speed Here the two distances are given, so find the time for each part. First part:

\frac{200}{100} = 2

hours Second part:

\frac{150}{75} = 2

hours Total distance =200+150=350 km and total time =2+2=4 hours Average speed

= \frac{350}{4} = 87.5

km/h Hence the correct option is (b).

The mean proportional between 36 and a number N is three times the mean proportional between 8 and 32. The number N is:

A.47
B.58
C.51
D.64

Solution

Mean proportional The mean proportional between a and b is

\sqrt{a b}

. Between 8 and 32:

\sqrt{8 \times 32} = \sqrt{256} = 16

Three times this value

= 3 \times 16 = 48

So the mean proportional between 36 and N is 48:

\sqrt{36 N} = 48

. Squaring both sides: 36N=2304

N = \frac{2304}{36} = 64

Hence the correct option is (d).

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