**Dear Aspirants,** **Our SSC Crackers team** is providing a new series of Quantitative Aptitude Questions for **Upcoming Exam **so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

**1) An article is sold at a profit of 25%. If the same article was sold at a loss of 30%, the selling price then would have been Rs. 572 lower than the current selling price. Find the current selling price of the article.**

a) Rs. 1040

b) Rs. 1300

c) Rs. 728

d) Rs. 1240

**Directions (2 – 5): Study the following graph and answer accordingly. The following graph shows the number of employees recruited (in thousands) in three banks SBI, ICICI and HDFC over the years from 2002 to 2007.**

**2) Total number of employees in ICICI Bank and HDFC bank together in the year 2004 was approximately what percent of ICICI and HDFC bank together in year 2007?**

a) 85

b) 80

c) 75

d) 90

**3) What is the approximate average number of employees in SBI over all the years together?**

a) 1990

b) 2085

c) 2300

d) 1800

**4) What is the approximate % increase in intake of ICICI bank from years 2003-2006 to 2004-2007?**

a) 27.43%

b) 27.77%

c) 27.91%

d) 27.16%

**5) Out of 3 banks, which bank succeeded in recruiting the most number of employees during the given years?**

a) SBI

b) ICICI

c) HDFC

d) Cannot be determined

**6) If A + B = 90, what is the simplified value of 1 + tan A cos B cosec B?**

a) sin^{2}A

b) sec^{2}A

c) cos^{2}A

d) tan^{2}A

**7) If cot (µ) = 2/3, find the value of (-5sin ^{2 }(µ) + 4cos (2µ) + 5).**

a) 1

b) 0

c) 2

d) 3

**8) If (cosecA + sinA) = 3/√2, then what is the value of (cosecA – sinA)?**

a) √2

b) 1/2

c) 1/√2

d) 2

**9) What is the simplified value for [{(cot(α)) ^{2} / (1 + (cot(α))^{2})} – {(tan(α))^{2} / (1 + (tan(α))^{2})}]?**

a) 1

b) cos (2α)

c) cos (α)

d) sin (2α)

**10) If x = (cosecθ – cosθ) and y = (sinθ – secθ), then what is the value of (x/y)?**

a) -tanθ

b) -cotθ

c) tanθ

d) cotθ

**Answers :**

**1) Answer: B **

Let the cost price of the article be Rs. x.

Then, x + 25% of x – (x – 30% of x) = 572

On solving the above equation, we get x = Rs. 1040

Thus, the current selling price of the article = 1040 + 0.25 × 1040 = Rs. 1300

**2) Answer: A **

In the year 2004, employees in ICICI +HDFC bank = (2.5+3) = 5.5 X 1000 = 5500

In the year 2007, employees in ICICI +HDFC bank = (3.5+3) = 6.5 X 1000 = 6500

Required percentage = 5500/6500 x 100 = 84.61% » 85%.

**3) Answer: B **

Required average = (1+2+1.5+2.5+3+2.5)/6

= (12.5/6) * 1000

= 2083.33

**4) Answer: B**

Employee intake of ICICI bank during 2003-2006 = (1+2.5+3+2.5) x 1000 = 9 x 1000 = 9000

Employee intake of ICICI bank during 2004-2007 = (2.5+3+2.5+3.5) x 1000 = 11500

% increase = (2500/9000)x 100 = 27.77%

**5) Answer: C **

No. of employees recruited by SBI = (1+2+1.5+2.5+3+2.5) x 1000 = 12500

No. of employees recruited by ICICI Bank = (1.5+1+2.5+3+2.5+3.5) x 1000 = 14 x 1000 = 14000

No. of employees recruited by HDFC bank = (0.5+2.5+3+3.5+3.5+3)x1000

16×1000 =16000.

Clearly, HDFC bank recruited maximum number of employees.

**6) Answer: B **

A + B = 90

∴ cos B = sin A and cosec B = sec A

∴ 1 + tan A cos B cosec B

= 1 + tan A sin A sec A

= 1 + (sin A/cos A) X (sin A) X (1/cos A)

= 1 + (sin^{2}A/cos^{2}A)

= (cos^{2}A + sin^{2}A)/cos^{2}A

= sec^{2}A

**7) Answer: B **

If cot (µ) = 2/3,

Cos (µ) / sin (µ) = 2/3

Cos (µ) = 2sin (µ) / 3

cos^{2 }(µ) = 4sin^{2 }(µ) / 9

9 cos^{2 }(µ) – 4sin^{2 }(µ) = 0

4cos^{2 }(µ) – 4sin^{2 }(µ) + 5 cos^{2 }(µ) = 0

4cos (2µ) + 5(1 – sin^{2 }(µ)) =0

Thus, 4 cos (2µ) + 5 – 5 sin^{2 }(µ) =0

**8) Answer: C**

(cosecA + sinA) = 3/√2

(cosecA + sinA)^{2} = (3/√2)^{2}

(cosec^{2}A + sin^{2}A + 2cosecA.sinA) = 9/2

[cosec^{2}A + sin^{2}A + 2.(1/sinA).sinA] = 9/2

(cosec^{2}A + sin^{2}A + 2) = 9/2

(cosec^{2}A + sin^{2}A) = (9/2) – 2

(cosec^{2}A + sin^{2}A) = (5/2) …. (1)

(cosecA – sinA)^{2} = (cosec^{2}A + sin^{2}A – 2cosecA.sinA)

From (1)-

(cosecA – sinA)^{2 }= (5/2) – 2 * (1/sinA) * (sinA)

(cosecA – sinA)^{2 }= (5/2) – 2

(cosecA – sinA)^{2 }= 1/2

(cosecA – sinA) = (1/√2)

**9) Answer: B **

The given expression can be simplified as follows-

[{((cos(α))^{2} / (sin(α))^{2}) / (cosec(α))^{2}} – {((sin(α))^{2} / (cos(α))^{2}) / (sec(α))^{2} }

[(cos(α))^{2} – (sin(α))^{2}] = cos(2α)

**10) Answer: B **

x = (cosecθ – cosθ) and y = (sinθ – secθ)

(x/y) = (cosecθ – cosθ)/(sinθ – secθ)

(x/y) = [(1/sinθ) – cosθ]/[sinθ – (1/cosθ)]

(x/y) = [(1 – sinθcosθ)/sinθ]/[(sinθcosθ – 1)/cosθ]

(x/y) = -cosθ/sinθ

(x/y) = -cotθ