##### Daily Quiz / QUANTS PRACTICE TESTS / Banking and SSC/RRB 2022 - Quants Practice Test -22.09.2022

#### Banking and SSC/RRB 2022 - Quants Practice Test -22.09.2022

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Q.1) Directions (Q. 1 – 5) : Find out the missing number in the following number series.

24, ?, 38, 87, 284, 1295

24 × 1 – 1

23 × 2 – 2

38 × 3 – 3

87 × 4 – 4

284 × 5 – 5

^{3}=**23**23 × 2 – 2

^{3}= 3838 × 3 – 3

^{3}= 8787 × 4 – 4

^{3}= 284284 × 5 – 5

^{3}= 1295###### Q.2) 64, 96, ?, 216, 324, 486

64 × 1.5 = 96

96 × 1.5 =

144 × 1.5 = 216

216 × 1.5 = 324

324 × 1.5 = 486

96 × 1.5 =

**144**144 × 1.5 = 216

216 × 1.5 = 324

324 × 1.5 = 486

###### Q.3) 122, 140, ?, 192, 228, 272, 325

The difference of difference is, 5, 6, 7, 8, …..

The answer is 163.

###### Q.4) ?, 368, 358, 330, 248, 4

**372**– (3

^{1}+1) = 368

368 – (3

^{2}+1) = 358

358 – (3

^{3}+1) = 330

330 – (3

^{4}+1) = 248

248 – (3

^{5}+1) = 4

###### Q.5) 48, 24, ?, 48, 192, 1536, 24576

48 × 1/2 = 24

24 × 1 =

24 × 2 = 48

48 × 4 = 192

192 × 8 = 1536

1536 × 16 = 24576

24 × 1 =

**24**24 × 2 = 48

48 × 4 = 192

192 × 8 = 1536

1536 × 16 = 24576

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Q.6) The least value of 8 cosec^{2} θ + 25 sin^{2} θ is:

Given: 8 cosec

Formula : If a cosec

Calculation:

Minimum Value of 8 cosec

Þ 2√200 Þ 2√(100 × 2)

Þ 2 × 10√2

Þ 20√2

^{2}θ + 25 sin^{2}θFormula : If a cosec

^{2}θ + b sin^{2}θ, then minimum value = 2√abCalculation:

Minimum Value of 8 cosec

^{2}θ + 25 sin^{2}θ = 2√(8 × 25)Þ 2√200 Þ 2√(100 × 2)

Þ 2 × 10√2

Þ 20√2

###### Q.7) The diagonal and length of the rectangle is 15 cm and 12 cm respectively. Find the area of the rectangle.

By Pythagoras theorem,

AC

^{2}= AB

^{2}= BC

^{2}

BC

^{2}= AC

^{2}– AB

^{2}

= 15

^{2}– 12

^{2}

= 225 – 144

BC

^{2}= 81

BC = 9 cm = breadth

Area of the rectangle = ℓ × b sq. units

= 12 × 9 cm

^{2}

= 108 cm

^{2}

###### Q.8) ₹6300 is divided between A, B, C such that A : B = 7 : 5, B : C = 4 : 3. Find the share of B.

A : B = 7 : 5 … (1)

B : C = 4 : 3 … (2)

(1) × 4 (2) × (5)

A : B = 28 : 20

B : C = 20 : 15

A : B : C = 28 : 20 : 15

Let A = 28x, B = 20x, C = 15x

28x + 20x + 15x = 6300

63x = 6300

x = 100

Share of B = 20x

= 20 (100)

B = 2000

B : C = 4 : 3 … (2)

(1) × 4 (2) × (5)

A : B = 28 : 20

B : C = 20 : 15

A : B : C = 28 : 20 : 15

Let A = 28x, B = 20x, C = 15x

28x + 20x + 15x = 6300

63x = 6300

x = 100

Share of B = 20x

= 20 (100)

B = 2000

###### Q.9) 15 men can complete a task in 10 days. In how many days can 20 men complete the same task?

We know that

M

15 × 10 = 20 × D

D

D

M

_{1}× D_{1}= M_{2}× D_{2}15 × 10 = 20 × D

_{2}D

_{2}= (15 × 10)/20D

_{2}= 7.5 days###### Q.10) Richa travels from A to B at the speed of 15 km/h, from B to C at 20 km/h, and from C to D at 30 km/h. If AB = BC = CD, then find the Richa's average speed.

Richa travels from A to B at the speed = 15 km/hr

Richa travels from B to C at the speed = 20 km/hr

Richa travels from C to D at the speed = 30 km/hr

AB = BC = CD

Formula:

Average speed = Total distance/Total time

Calculation:

Let AB = BC = CD = 60 km (LCM of 15, 20 and 30)

Total distance = 3 × 60 = 180 km

Time taken by Richa to travels from A to B = 60/15 = 4 hrs.

Time taken by Richa to travels from B to C = 60/20 = 3 hrs.

Time taken by Richa to travels from C to D = 60/30 = 2 hrs.

Total time taken by Richa to travels from A to D

= 4 + 3 + 2 = 9 hrs.

Average speed = 180/9 = 20 km/hr.

Richa travels from B to C at the speed = 20 km/hr

Richa travels from C to D at the speed = 30 km/hr

AB = BC = CD

Formula:

Average speed = Total distance/Total time

Calculation:

Let AB = BC = CD = 60 km (LCM of 15, 20 and 30)

Total distance = 3 × 60 = 180 km

Time taken by Richa to travels from A to B = 60/15 = 4 hrs.

Time taken by Richa to travels from B to C = 60/20 = 3 hrs.

Time taken by Richa to travels from C to D = 60/30 = 2 hrs.

Total time taken by Richa to travels from A to D

= 4 + 3 + 2 = 9 hrs.

Average speed = 180/9 = 20 km/hr.