First we should know some mathematical operations. They are add (+), subtraction (–), multiply (×) and division (÷), greater than (>), less than (<). This test is set up to test candidates skill in mathematical operations. The questions involving these operations are set using artificial symbols. You are required to substitute the real signs and solve the questions accordingly, to get the answer.
There are three types of questions based on mathematical operations which are asked in various competitive examinations. They are,
In such type of questions you have some substitutes for various mathematical symbols or numerals followed by a question involving calculation of an expression or choosing the correct/incorrect equation.
Brackets
Of
Division
Multiplication
Addition
Subtraction
While solving a mathematical operations proceed according to the BODMAS’ formula.
Example 1: If ‘+’ means ‘minus’ ‘×’ means ‘divided by’ ‘÷’ means ‘plus’ and ‘–’means ‘multiplied by’ then which of the following will be the value of expression 7 × 3.5 ÷ 2 – 4 + 5?
(a) 4
(b) 5
(c) 11
(d) None of these
Correct Answer is Option (b)
Using the proper notations in the given expression, we have = 7 × 3.5 ÷ 2 – 4 + 5 = 7 + 3.5 + 2 ÷ 4 – 5 = 2 + 2 × 4 – 5 = 2 + 8 – 5 = 10 – 5 = 5.
Example 2: If × means +, + means ÷, – means × and ÷ means –, then 6 × 4 – 5 + 2 ÷ 1 =?
(a) 10
(b) 11
(c) 12
(d) 15
Correct Answer is Option (d)
Using the proper notations in the given expression, we have 6 × 4 – 5 + 2 ÷ 1 = 6 + 4 × 5 ÷ 2 – 1 = 6 + 4 × 2.5 – 1 = 6 + 10 – 1 = 16 – 1 = 15.
Example 3: If P denotes ‘multiplied by’ T denotes ‘subtracted from, M denotes ‘added to’ and B denotes ‘divided by’, then 12 P 6 M 15 T 16 B 4
(a) 70
(b) 83
(c) 75
(d) 110
(e) None of these
Correct Answer is Option (b)
12 P 6 M 15 T 16 B 4 = 12 × 6 + 15 – 16 ÷ 4 = 12 × 6 + 15 – 4 = 72 + 15 – 4 = 87 – 4 = 83
This type of question certain signs or numbers interchanging with each other. The candidate is required to change the given signs or change the given numbers with each other and select which of the equation is correct of the given alternatives.
Example 4: If signs + and –and numbers 4 and 8 interchanges with each other, which one of the following four equations would be correct?
(a) 4 – 8 + 12 = 0
(b) 8 – 4 ÷ 12 = 8
(c) 4 ÷ 8 – 12 = 16
(d) 8 ÷ 4 – 12 = 24
Correct Answer is Option (a)
On interchanging signs + and – and numbers 4 and 8 in equation
(a) 8 + 4 – 12 = 0
⇒ 12 – 12 = 0
⇒ 0 = 0
Example 5: Which one of the four interchanges in signs and number would make the given equation correct? 6 × 4 + 2 = 16
(a) + and ×, 2 and 4
(b) + and ×, 4 and 6
(c) + and ×, 2 and 6
(d) None of the above
Correct Answer is Option (b)
On interchanging signs + and × and 4 and 6, 4 + 6 × 2 = 4 + 12 = 16
Example 6: If 5 × 4= 15, 7 × 8 = 49 and 6 × 5 = 24, then 8 × 4 =?
(a) 24
(b) 26
(c) 28
(d) 30
Correct Answer is Option (a)
As, 5 × 4 = 5 × (4 – 1) = 5 × 3 = 15, 7 × 8 = 7 × (8 – 1) = 7 × 7 = 49 and 6 × 5 = 6 × (5 – 1) = 6 × 4 = 24 Similarly, 8 × 4 = 8 × (4 –1) = 8 × 3 = 24
Example 7: If 64 × 52 = 17, 48 × 56 = 23 and 74 × 35 = 19 then 84 × 37 =?
(a) 32
(b) 28
(c) 22
(d) 20
As, 64 × 52
⇒ (6 + 4) + (5 + 2) = 17, 48 × 56
⇒ (4 + 8) + (5 + 6) = 23 and 74 × 35
⇒ (7 + 4) + (3 + 5) = 19
Similarly, 84 × 37 Þ (8 + 4) + (3 + 7) = 22
In this type of questions certain relations between different sets of elements is given (in terms of ‘less than’, ‘greater than’, or ‘equal to’), using either the real symbols or substituted symbols. The candidate is required to briefly read the given statements and then choose which of the conclusions is/are definitely true. Directions (Examples 8 to 10) In the following questions, the symbols d, @, ©, % and * are used with the following means as illustrated below:
Now in each of the following questions assuming the given statements to be true, find which of the three conclusions I, II, III and IV given below them is/are definitely true and give your answer accordingly.
Example 8: Statements D d T, T @ R, R © M, M % K Conclusions I. R @ D II. R % D III. K « T IV. M d T
(a) Only either I or II is true
(b) Only III and IV are true
(c) Only either I or II and III are true
(d) Only either I or II and III and IV are true
Correct Answer is Option (a)
Here, D d T
⇒ D £ T; T @ R
⇒ T = R; R ©M
⇒ R ^{3 }M; M % K
⇒ M > K So, D £ T = R^{ 3} M > K
Now, R @ D
⇒ R = D (False); R % D
⇒ R > D (False) K « T
⇒ K < T (True) M d T
⇒ M £ T (True) hence, only either I or II and III and IV are true.
Example 9: Statements J @ F, F d N, N % H, H © G Conclusions I. G « N II. N ©J III. F « J IV. J d G
(a) Only I and II are true
(b) Only I, II and III are true
(c) Only I, III and IV are true
(d) All I, II, III and IV are true
Correct Answer is Option (a)
Here, J @ F
⇒ J = F; F d N = F £ N N %H
⇒ N > H; H © G = H^{3} G So, J = F £ N >H^{3} G
Now, G «N
⇒ G < N (True); N © J
⇒ N^{3} J (True) F « J
⇒ F < J (False): J d G
⇒ J £ G (False) Hence, only I and II are true.
Example 10: Statements R « K, K % D, D @ V, V d M Conclusions I. R « D II. V « R III. D @ M IV. M % D
(a) None is true
(b) Only III is true
(c) Only IV is true
(d) Only either III or IV is true
Correct Answer is Option (d)
Here, R « K
⇒ R < K; K % D
⇒ K > D D @ V
⇒ D = V; V d M = V £ M
So, R < K > D = V £ M
Now, R « D
⇒ R < D (False); V « R
⇒ S V< R (False) D @M
⇒ D ?? M (False); M % D
⇒ M > D (False) But either III or IV is true.
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