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SSC GD Constable 12 February 2025 · Shift 3 · Elementary Mathematics — questions with answers · ShikshaSphere

Chapter 103 of 210

12 February 2025 · Shift 3

Elementary Mathematics

20 questions ~20 min readFree

20 questions

An aeroplane travels distances

1320 km, 1760 km

and 2095 km at the rate of

528 km / h, 352 km / h

and 419

km / h

, respectively. Find the average speed (in km/h) of the aeroplane.

A.398
B.497
C.452
D.414

Solution

Average speed Average speed = total distance ÷ total time, so the time of each leg is needed first.

t_{1} = \frac{1320}{528} = 2.5

t_{2} = \frac{1760}{352} = 5

t_{3} = \frac{2095}{419} = 5

h. Total distance =1320+1760+2095=5175 km. Total time =2.5+5+5=12.5 h. Average speed

= \frac{5175}{12.5} = 414

km/h. Hence option (d) is correct.

Find the mean proportional between

(4 + \sqrt{5})

and

(12 - 2 \sqrt{20})

A. $\sqrt{39}$
B. $2 \sqrt{7 - \sqrt{5}}$
C. $\sqrt{7 - \sqrt{5}}$
D. $2 \sqrt{7 + \sqrt{5}}$

Solution

Mean proportional The mean proportional between a and b is

\sqrt{a b}

. First simplify the surd:

12 - 2 \sqrt{20} = 12 - 2 \times 2 \sqrt{5} = 12 - 4 \sqrt{5}

a b = (4 + \sqrt{5}) (12 - 4 \sqrt{5}) = 48 - 16 \sqrt{5} + 12 \sqrt{5} - 4 \times 5

a b = 48 - 20 - 4 \sqrt{5} = 28 - 4 \sqrt{5}

\sqrt{a b} = \sqrt{4 (7 - \sqrt{5})} = 2 \sqrt{7 - \sqrt{5}}

Hence option (b) is correct; option (c) is the same surd without the factor 2 taken out of

\sqrt{4}

Ayush starts his journey from Chandigarh to Ludhiana with his car at a speed of

77 km / hr

and returns at a speed of

63 km / hr

. If the distance between Chandigarh and Ludhiana is 97 km , then what is the average speed (in km/hr) of Ayush during his whole journey?

A.69
B.68.2
C.68
D.69.3

Solution

Average speed both ways The two halves of the journey are of EQUAL distance, so the average speed is the harmonic mean, not

\frac{77 + 63}{2}

. Formula: average speed

= \frac{2 x y}{x + y}

; the 97 km is not needed at all.

= \frac{2 \times 77 \times 63}{77 + 63} = \frac{9702}{140}

=69.3 km/hr. Hence option (d) is correct; option (a) 69 is the trap of taking

\frac{77 + 63}{2} = 70

and rounding.

What is the smallest number which is when increased by 5 is divisible by 18 , 30, 36 and 24 ?

A.355
B.365
C.350
D.360

Solution

LCM application The number becomes divisible by all four only when (number

+ 5

) is a common multiple of them, and the smallest such multiple is their LCM.

18 = 2 \times 3^{2}

30 = 2 \times 3 \times 5

36 = 2^{2} \times 3^{2}

24 = 2^{3} \times 3

. LCM

= 2^{3} \times 3^{2} \times 5 = 360

. Required number =360-5=355. Hence option (a) is correct; 360 (option d) is the LCM itself, not the number asked for.

Two trains are moving in opposite directions at speeds of

80 km / h

and 70

km / h

. The length of one train is 180 m . The time taken by them to cross each other is 19 seconds. The length (in m ) of the other train, correct to 2 decimal places, is:

A.614.03
B.611.66
C.610.21
D.613.04

Solution

Trains crossing each other They move in opposite directions, so the speeds add: relative speed =80+70=150 km/h.

150 km/h

= 150 \times \frac{5}{18} = \frac{125}{3}

m/s. Crossing each other means covering the SUM of the two lengths. Sum of lengths

= \frac{125}{3} \times 19 = \frac{2375}{3}

m. Other train

= \frac{2375}{3} - 180 = \frac{2375 - 540}{3} = \frac{1835}{3} = 611.66 \dots

m. Hence option (b) is correct.

p

varies directly to

q

and

p_{1} = 15

q_{1} = 630

, and

p_{2} = 6

, then find the value of

q_{2}

A.7
B.84
C.42
D.252

Solution

Direct variation p varies directly as q means

\frac{p}{q}

stays constant, so

\frac{p_{1}}{q_{1}} = \frac{p_{2}}{q_{2}}

\frac{15}{630} = \frac{6}{q_{2}}

q_{2} = \frac{6 \times 630}{15} = 6 \times 42 = 252

Hence option (d) is correct; 42 (option c) is only

\frac{630}{15}

, the constant, not

q_{2}

A and B working separately can finish a work in 18 days and 24 days, respectively. A begins the work and works for two consecutive days, while B works on the

3^{rd}

day. This pattern is continued till the work is completed. In how many days will the work be completed?

A. $14 \frac{3}{5}$
B. $16 \frac{2}{3}$
C. $23 \frac{2}{3}$
D. $19 \frac{1}{2}$

Solution

Time and work in cycles Let the total work = LCM(18,24)=72 units. A does

\frac{72}{18} = 4

units/day and B does

\frac{72}{24} = 3

units/day. One cycle is A, A, B, i.e. 4+4+3=11 units in 3 days.

6 cycles =18 days give

6 \times 11 = 66

units; 72-66=6 units are left. Day 19 belongs to A: he does 4 units, so 6-4=2 units are still left. Day 20 is again A's day, and he needs only

\frac{2}{4} = \frac{1}{2}

day for the rest. Total time

= 18 + 1 + \frac{1}{2} = 19 \frac{1}{2}

days. Hence option (d) is correct.

The salary of a worker is first increased by 8% and then it is decreased by 8%. What is the percentage change in his salary ?

A.0.81%
B.0.25%
C.0.49%
D.0.64%

Solution

Successive percentage change A rise of x% followed by a fall of the same x% never cancels out; the net effect is always a fall of

\frac{x^{2}}{100}

%. Here x=8, so net change

= - \frac{8 \times 8}{100} = - 0.64

%. Check with ₹100:

100 \times \frac{108}{100} = 108

, then

108 \times \frac{92}{100} = 99.36

. The fall is ₹0.64 on ₹100, i.e. 0.64%. Hence option (d) is correct.

David has two grandsons Rodney and Amit. 15 year old Rodney gets some money from David's wealth and 16 year old Amit gets rest of the money. But Rodney and Amit will get money only when they turn 22 years old. Till then the money is in a bank getting interest at rate 8% compounded annually. When both turn 22, they receive the same amount. How much had David given Amit (in ₹) initially, if total money with David was ₹23400?

A.12150
B.12500
C.11000
D.11250

Solution

Compound interest shares Rodney is 15, so his money grows for 22-15=7 years; Amit is 16, so his grows for 22-16=6 years. At 8% per annum the yearly growth factor is

1 + \frac{8}{100} = \frac{27}{25}

Equal maturity amounts:

R \times {(\frac{27}{25})}^{7} = A \times {(\frac{27}{25})}^{6}

\frac{R}{A} = \frac{25}{27}

, so the two shares are in the ratio 25:27. (25+27)=52 units = ₹23,400, so 1 unit

= \frac{23400}{52} =

₹450. Amit's share

= 27 \times 450 =

₹12,150. Hence option (a) is correct; ₹11,250 is Rodney's share

(25 \times 450)

A profit earned when an item is sold for ₹4,000 is ten times the loss incurred when it is sold for ₹2,790. At what price should the item be sold if it is desired to make a profit of 20%?

A.₹3,075
B.₹3,860
C.₹3,480
D.₹2,900

Solution

Profit and loss equation Let the cost price be ₹x. Profit at ₹4,000 =4000-x; loss at ₹2,790 =x-2790. Given: 4000-x=10(x-2790) 4000-x=10x-27900

11 x = 31900 \Rightarrow x = 2900

For a 20% profit, SP

= 2900 \times \frac{120}{100} =

₹3,480. Hence option (c) is correct; ₹2,900 (option d) is only the cost price.

A merchant fixes the marked price of his goods at 60% above the cost price. He sells his goods at 25% discount. His percentage of profit is:

A.25%
B.15%
C.10%
D.20%

Solution

Marked price and discount Let the cost price = ₹100 (percentages are all relative, so any convenient CP works). Marked price =100+60= ₹160. Discount of 25%: SP

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

₹120. Profit =120-100= ₹20 on a cost of ₹100, i.e. 20%. Hence option (d) is correct; 60-25=35 is not the answer, because the discount acts on ₹160, not on ₹100.

A square field has its area equal to

252 m^{2}

. Find its perimeter.

A. $25 \sqrt{7} m$
B. $23 \sqrt{7} m$
C. $24 \sqrt{7} m$
D. $22 \sqrt{7} m$

Solution

Area and perimeter of a square For a square, area

= {(side)}^{2}

, so side

= \sqrt{area}

. Side

= \sqrt{252} = \sqrt{36 \times 7} = 6 \sqrt{7}

Perimeter

= 4 \times side = 4 \times 6 \sqrt{7}

Perimeter

= 24 \sqrt{7}

m. Hence option (c) is correct.

What will be difference in population 3 years ago and 2 years ago of a town, whose current population is 100000 and which is increasing at a rate of 25% every year?

A.12050
B.12800
C.13150
D.12250

Solution

Population growth backwards A 25% yearly rise means each year's population is

\frac{5}{4}

of the previous year's, so going BACK means dividing by

\frac{5}{4}

. 2 years ago

= 100000 \times {(\frac{4}{5})}^{2} = 100000 \times \frac{16}{25} = 64000

. 3 years ago

= 64000 \times \frac{4}{5} = 51200

Required difference =64000-51200=12800. Hence option (b) is correct; subtracting 25% of 100000 twice gives the wrong figures, because the rate applies to the earlier year, not to the present one.

Evaluate:

(- 9) - (- 60) \div (- 15) + (- 3) \times 9

A.- 43
B.- 42
C.- 40
D.-39

Solution

BODMAS with integers By BODMAS, division and multiplication are done before addition and subtraction. Division:

(- 60) \div (- 15) = + 4

(a negative divided by a negative is positive). Multiplication:

(- 3) \times 9 = - 27

. Now the expression is (-9)-(+4)+(-27) =-9-4-27=-40 Hence option (c) is correct; -40 turns into -42 only if the

\div

sign is mistakenly given a negative value.

The simple interest on a principal amount (in ₹) is ₹191 for a period of 5 years at the rate of 2% per annum. The principal amount (in ₹) is:

A.1910
B.1908
C.1914
D.1913

Solution

Simple interest principal Formula: SI

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

191 = \frac{P \times 2 \times 5}{100} = \frac{10 P}{100} = \frac{P}{10}

P = 191 \times 10 = 1910

So the principal is ₹1,910. Hence option (a) is correct.

The population of a district is 333000, out of which 213000 are males. 71% of the population is literate. If 71% males are literate, then what percentage of females are literate?

A.68%
B.69%
C.74%
D.71%

Solution

Percentage of literacy Females =333000-213000=120000. Literate people =71% of

333000 = \frac{71}{100} \times 333000 = 236430

. Literate males =71% of

213000 = \frac{71}{100} \times 213000 = 151230

. Literate females =236430-151230=85200.

Required percentage

= \frac{85200}{120000} \times 100 = 71

. Hence option (d) is correct: when the overall rate equals the male rate, the female rate must equal it too.

The LCM of

2^{3} \times 9^{2} \times 13, 2^{2} \times 13^{2} \times 19

and

9^{3} \times 13^{2} \times 19^{2}

is:

A. $2^{3} \times 9^{3} \times 13^{2} \times 19^{2}$
B. $2^{2} \times 9^{3} \times 13 \times 19^{2}$
C. $2^{2} \times 9^{2} \times 13^{2} \times 19^{3}$
D. $2^{3} \times 9^{2} \times 13^{2} \times 19$

Solution

LCM of factorised numbers The LCM takes the HIGHEST power of every base that appears in any of the numbers (the HCF would take the lowest).

Powers of 2 present:

2^{3}

and

2^{2}

- take

2^{3}

. Powers of 9 present:

9^{2}

and

9^{3}

- take

9^{3}

. Powers of 13 present: 13,

13^{2}

13^{2}

- take

13^{2}

; powers of 19: 19,

19^{2}

- take

19^{2}

. LCM

= 2^{3} \times 9^{3} \times 13^{2} \times 19^{2}

. Hence option (a) is correct.

There are 30 members in group

A

, 76 members in group B and 44 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 ₹249, ₹168 and ₹498, respectively. The total average amount (in ₹) spent per member is:

A.284
B.282
C.281
D.279

Solution

Weighted average The overall average is total amount ÷ total members, never the plain average of ₹249, ₹168 and ₹498. Group A:

30 \times 249 = 7470

. Group B:

76 \times 168 = 12768

. Group C:

44 \times 498 = 21912

Total amount =7470+12768+21912=42150; total members =30+76+44=150. Average per member

= \frac{42150}{150} = 281

. Hence option (c) is correct; the plain average

\frac{249 + 168 + 498}{3} = 305

ignores the group sizes.

A dishonest shopkeeper promises to sell his goods at cost price. However, he uses a weight that actually weighs 37% less than what is written on it. Find his profit percentage.

A.60 92/63%
B.59 47/63%
C.57 46/63%
D.58 46/63%

Solution

Profit by false weight The weight stamped 100 g actually gives only 100-37=63 g of goods, but the customer pays for 100 g. So the shopkeeper's cost is for 63 units while he charges for 100 units. Profit %

= \frac{gain}{cost} \times 100 = \frac{100 - 63}{63} \times 100 = \frac{3700}{63}

63 \times 58 = 3654

and 3700-3654=46, so

\frac{3700}{63} = 58 \frac{46}{63}

. Hence option (d) is correct; dividing by 100 instead of 63 would wrongly give 37%.

The value of 40 − 3 × [10 + 6 × {20 − 10(6 − 5) × 2} ÷ 47] is:

A.7
B.5
C.1
D.10

Solution

BODMAS with nested brackets Work outward from the innermost bracket: (6-5)=1. Inside the braces:

10 \times 1 \times 2 = 20

, so

{20 - 20} = 0

. Inside the square bracket:

6 \times 0 \div 47 = 0

, so [10+0]=10.

40 - 3 \times 10 = 40 - 30 = 10

Hence option (d) is correct; the whole brace collapses to zero, which is what makes the ÷ 47 harmless.

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