Saturday, February 8, 2014

Important Questions of Quantitative Aptitude for SSC CGL Tier I-2014

Posted by Jasleen Behl
In my previous session, i discussed some important questions of General Intelligence and Reasoning for upcoming SSC CGL Tier I exam. Today, I'll discuss questions of Quantitative Aptitude section which will give you an outlook of exam pattern. As we all know, this section also includes data interpretation, I will discuss its questions in parts.

Quantitative Aptitude

1. If  `S_1` = 2, `S_2` = 3 and `S_(n+3) = S_(n+1) + S_(n+2)`, then what will be the fifth term of the sequence?


Solution:  Sequence and Series
Given that `S_1` = 2, `S_2` = 3

To find: `S_5` 

Now, given pattern of sequence is  `S_(n+3) = S_(n+1) + S_(n+2)`

If n = 2,
 `S_(2+3) = S_(2+1) + S_(2+2)`
`S_5 = S_3 + S_4`

Similarly,
If n = 0,
`S_3 = S_1 + S_2` and
If n = 1,
`S_4 = S_2 + S_3`

Putting values of `S_3` and `S_4` in equation of `S_5`

`S_5 = S_3 + S_4`
⇒`S_5 = S_3 + S_2 + S_3`
⇒`S_5 =  2(S_1 + S_2) + S_2`
⇒`S_5 = 2(S_1) + 3(S_2)`
⇒`S_5 = 2(2) + 3(3)`
⇒`S_5 = 4+ 9` = `13`

2. A group of `12` men and `16` women can do a piece of work in `5` days and a group of `13` men and `24` women can do the same work in `4` days. How many days will `7` men and `10` women take to finish the same work?

Solution: Time and Work
  12 Men + 16 Women i.e. 12 M + 16 W takes 5 days
and 13 M + 24 W takes 4 days

I will solve this question using per day work,

In one day, work is done by : 5(12 M + 16 W) and
4(13 M + 24 W)

Therefore, 5(12 M+ 16 W) = 4(13 M + 24W)
⇒ 1M = 2W

So, equation becomes: 12M + 16 W = 40 W (takes 5 days)

To find: 7 M + 10 W = 24 W

Using basic formula, `M_1 times D_1 = M_2 times D_2`

`D_2 = (40 times 5)/ (24) = 8.33` days

3. Base of a right angled triangle is 8 cm and hypotenuse is 17 cm. Find the area of triangle.

Solution: Using Pythagoras theorem:

`(Hypotenuse)^2 = (base)^2 + (perpendicular)^2`

`(17)^2 = 8^2 + P^2`
⇒ `289 = 64 + P^2`
⇒P = 15 cm

Now area of right angled triangle = `1/2` `times` Base `times` Perpendicular

Area = `(1/2) times 8 times 15`
⇒ 60 `cm^2`

4. Rs 1100 is divided among A, B and C in such a way that the ratio of amounts of A and B is 3: 2 and that of B and C is 5: 4. Find the amounts each received?

Ratios are:
A: B = 3: 2
B: C = 5: 4

A: B: C = 15 : 10 : 8  

A's share = `(1100 times 15)/ (15+10+8)`
⇒`(1100 times 15)/ 33`
⇒Rs 500

B's share = `(1100 times 10)/ 33`
⇒ Rs 333.33
C's share = `(1100 times 8)/ 33`
⇒ Rs 266.66

5. If x is 5 more than y and sum of squares of x and y is 55, then what will be the product of x and y?

Solution: Given that: `x = y + 5`........................(1)

`x^2 + y^2 = 55` .................(2)
 To find: xy

We know that, `(x-y)^2 = x^2 + y^2 - 2xy`
From eqn 1 and 2

`5^2 = 55 - 2 xy`

⇒ 2xy = 30
⇒ xy = 15

6. In the following figure, Find angle D, where, AB = BC = CA = CD


Solution: Geometry
As sides AB = BC = CA
Therefore, angle ACB = ABC = BAC = `60^o` (sum of all angles = 180^o)

angle ACD = `180^o` - `60^o`
⇒ angle ACD = `120^o`

Now, as sides AC and CD ARE equal, 
Therefore, angle CAD = angle ADC

Also, CAD + ADC + ACD = `180^o`
So, angle D = `30^o`

7. If x + `1/x` = 2, then what will be the value of `x^2 + 1/(x^2)`?

Solution: Using formula, `(x+y)^2 = x^2 + y^2 + 2xy`

Here, y = 1/x
Therefore, `(x + (1/x))^2` = `x^2` + `1/(x^2)` +`2 (x)(1/x)`

⇒  `x^2 + 1/(x^2)` = `2^2` - 2 = 2

8. What will be the volume of right circular cylinder if radius = 2 cm and height = 2.5 cm?

Solution: Volume of right circular cylinder = `pi r^2 h`

r = 2 cm
h = 2.5 cm

Therefore, Volume = `(3.14) times 4 times (2.5)`
⇒ 31.4 cm

Soon i will update more questions for practice.
For General awareness and Reasoning, Check following article:

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