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  #1  
June 9th, 2015, 09:10 AM
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SSC CGL Arithmetic Material

Hello I am going to appear in Staff selection commission (SSC) combined graduate level (CGL) Exam but I have no study material for Arithmetic Ability . Would you please provide me a link from where I can download study material for any topic of Arithmetic Ability ??????
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  #2  
March 3rd, 2017, 08:05 AM
Unregistered
Guest User
 
Re: SSC CGL Arithmetic Material

Hi buddy here I am looking for SSC CGL exam Arithmetic subject study Material, so would you plz provide me sample question paper , if you have this ??
  #3  
March 3rd, 2017, 08:05 AM
Super Moderator
 
Join Date: Mar 2012
Re: SSC CGL Arithmetic Material

As you are asking for SSC CGL exam Arithmetic subject sample paper so on your demand I am providing same for you :


1. Find the HCF of 2.1, 1.05 and 0.63

0.44
0.64
0.21
None of above

Answer :

Option C

Explanation:

To solve this question quickly, first remove decimal by multiplying each term with 100,
Then terms become 210, 105, 63
Then HCF of above terms is 21,
So Answer is 0.21
2. Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest terms.

\begin{aligned} \frac{30}{25} \end{aligned}
\begin{aligned} \frac{28}{29} \end{aligned}
\begin{aligned} \frac{28}{29} \end{aligned}
\begin{aligned} \frac{16}{25} \end{aligned}

Answer :

Option D

Explanation:

We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25
3. Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest terms.

\begin{aligned} \frac{11}{12} \end{aligned}
\begin{aligned} \frac{23}{24} \end{aligned}
\begin{aligned} \frac{26}{27} \end{aligned}
\begin{aligned} \frac{4}{7} \end{aligned}

Answer :

Option A

Explanation:

HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12
4. Find the HCF of 54, 288, 360

18
36
54
108

Answer :

Option A

Explanation:

Lets solve this question by factorization method.

\begin{aligned}
18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
\end{aligned}

So HCF will be minimum term present in all three, i.e.
\begin{aligned}
2 \times 3^2 = 18
\end{aligned}
5. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is

\begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
\begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
\begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
\begin{aligned} 2 \times 3 \times 5 \end{aligned}

Answer :

Option C

Explanation:

As in HCF we will choose the minimum common factors among the given.. So answer will be third option
6. What will be the LCM of 8, 24, 36 and 54

54
108
216
432

Answer :

Option C

Explanation:

LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216
7. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is

10:28 am
10:30 am
10:31 am
None of above

Answer :

Option C

Explanation:

L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes

They will beep together at 10:31 a.m.

Sometimes questions on red lights blinking comes in exam, which can be solved in the same way


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