Question 11 Mark
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \text{B} \ \ \ \ 7\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ 6\ \ \ \ \text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ 9\ \ \ \ 8\ \ \ \ \ \text{A}}$
AnswerIf $\text{A}+\text{B}=8,\text{A}+\text{B}\geq9$ is possible only if $A = b = 9$ But from $7 + B = A, A = B = 9$ is impossible. Surely, $\text{A}+\text{B}=8, \text{A}+\text{B}\leq9$ So,$ A + 7 = 9,$ Surely $A = 27 + B = A, 7 + B = 2, B = 5 So, A = 2, B = 5$
View full question & answer→Question 21 Mark
Find the remainder, without performing actual division, when $798$ is divided by $11$.
AnswerLet $n = 798 = a$ multiple of $11 + [7 + 8 - 9] 798 = a$ multiple of $11 + 6$
$\therefore$ Remainder $= 6$
View full question & answer→Question 31 Mark
Which of the following statements are true? If a number divides three numbers exactly, it must divide their sum exactly.
Question 41 Mark
Without performing actual addition and division, write the quotient when the sum of $6$9 and $96.$ is divided by: $15$
AnswerTwo numbers are $69$ and $96$ whose digits are revesed Here $a = 6 = 9$ If it is divisible by $a + b$ i.e., $6 + 9 = 15$, then quotient $= 11$
View full question & answer→Question 51 Mark
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ 2 \ \ \ \ \text{A}\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ 6\ \ \ \ \text{A}\ \ \ \ \text{B}\ \ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \ 0\ \ \ \ \ 9}$
Answer$A + B = 9$ as the sum of two digits can never be $192 + A = 0, A$ must be $8A + 9, 8 + B = 9, B = 1$ So, $A = 8, B = 1$
View full question & answer→Question 61 Mark
Without performing actual addition and division, write the quotient when the sum of $69$ and $96.$ is divided by: $11$
AnswerTwo numbers are $69$ and $96$ whose digits are revesed Here $a = 6 = 9$ Sum if $69 + 96$ is divisible by $11,$ then quotient $= a + 6 = 6 + 9 = 15$
View full question & answer→Question 71 Mark
Which of the following statements are true? A number is divisible by $18,$ if it is divisible by both $3$ and $6.$
AnswerIt must be divisible by $9$ and $2$ both.
View full question & answer→Question 81 Mark
Without performing acute computations, find the quotient when $94 - 49$ is divisible by:
$5$
AnswerTwo numbers are $94$ and $49.$ Whose digits are reversed.When it is divided by $a - b$ i.e., $9 - 4 = 5,$ then quotient will be $= 9$
View full question & answer→Question 91 Mark
Which of the following statements are true? If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
AnswerFalse.Solution:
It is not necessarily.
View full question & answer→Question 101 Mark
Which of the following statements are true? If a number is divisible by $8,$ it must be divisible by $4.$
View full question & answer→Question 111 Mark
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \ 1\\ \ \ \ \ \underline{+\ \ \ 1\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \text{B}\ \ \ \ 0\ }$
View full question & answer→Question 121 Mark
Which of the following statements are true? If a number is divisible by $3,$ it must be divisible by $9.$
AnswerIt is not necessarily that divide by $9.$
View full question & answer→Question 131 Mark
Which of the following statements are true? If a number is divisible by both $9$ and $10,$ it must be divisible by $90.$
View full question & answer→Question 141 Mark
Without performing acute computations, find the quotient when $94 - 49$ is divisible by: $9$
AnswerTwo numbers are $94$ and $49.$ Whose digits are reversed. If $94 - 49$ is divided by $9,$ then the quotient $= a - b = 9 - 4 = 5$
View full question & answer→Question 151 Mark
Which of the following statements are true$?$ The sum of two consecutive odd numbers is always divisible by $4.$
View full question & answer→Question 161 Mark
Which of the following statements are true? If a number is divisible by $4,$ it must be divisible by $8$.
AnswerIt is not necessarily that it must divide by $8.$
View full question & answer→Question 171 Mark
Which of the following statements are true? If a number is divisible by $9,$ it must be divisible by $3.$
View full question & answer→Question 181 Mark
Which of the following statements are true? If two numbers are co-prime, at least one of them must be a prime number.
AnswerFalse.Solution:
It is not necessarily.
View full question & answer→Question 191 Mark
Solve Cryptarithms. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ \ \ \text{A} \ \ \ \ \text{B}\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ \text{A}\ \ \ \ \text{B}\ \ \ \ \ 1\ \ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \text{B}\ \ \ \ 0\ \ \ \ \ 1}$
Answer$B + 1 = 8, B = 7A + 1, A + 7 = 1, A = 4$ So, $A = 4, b = 7$
View full question & answer→Question 201 Mark
Without performing acute computations, find the quotient when 94 - 49 is divisible by:
9
AnswerTwo numbers are 94 and 49. Whose digits are reversed.
If 94 - 49 is divided by 9, then the quotient = a - b = 9 - 4 = 5
View full question & answer→Question 211 Mark
Without performing acute computations, find the quotient when 94 - 49 is divisible by:
5
AnswerTwo numbers are 94 and 49. Whose digits are reversed.
When it is divided by a - b i.e., 9 - 4 = 5, then quotient will be = 9
View full question & answer→Question 221 Mark
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by:
15
AnswerTwo numbers are 69 and 96 whose digits are revesed Here a = 6 = 9
If it is divisible by a + b i.e., 6 + 9 = 15, then quotient = 11
View full question & answer→Question 231 Mark
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by:
11
AnswerTwo numbers are 69 and 96 whose digits are revesed Here a = 6 = 9
Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
View full question & answer→Question 241 Mark
Solve Cryptarithms.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \text{B} \ \ \ \ 7\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ 6\ \ \ \ \text{A}\ \ \ \ \text{B}\ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ 9\ \ \ \ 8\ \ \ \ \ \text{A}}$
AnswerIf $\text{A}+\text{B}=8,\text{A}+\text{B}\geq9$ is possible only if A = b = 9 But from 7 + B = A, A = B = 9 is impossible. Surely, $\text{A}+\text{B}=8, \text{A}+\text{B}\leq9$
So, A + 7 = 9, Surely A = 27 + B = A, 7 + B = 2, B = 5
So, A = 2, B = 5
View full question & answer→Question 251 Mark
Solve Cryptarithms.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\ \ \ \ \text{A} \ \ \ \ \text{B}\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ \text{A}\ \ \ \ \text{B}\ \ \ \ \ 1\ \ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \text{B}\ \ \ \ 0\ \ \ \ \ 1}$
AnswerB + 1 = 8, B = 7A + 1, A + 7 = 1, A = 4
So, A = 4, b = 7
View full question & answer→Question 261 Mark
Solve Cryptarithms.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\ \ \ \ 2 \ \ \ \ \text{A}\ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \underline{+\ \ \ \ 6\ \ \ \ \text{A}\ \ \ \ \text{B}\ \ }\\\ \ \ \ \ {\ \ \ \ \ \ \ \ \ \ \ \ \ \text{A}\ \ \ \ 0\ \ \ \ \ 9}$
AnswerA + B = 9 as the sum of two digits can never be 192 + A = 0, A must be 8A + 9, 8 + B = 9, B = 1
So, A = 8, B = 1
View full question & answer→Question 271 Mark
Find the remainder, without performing actual division, when 798 is divided by 11.
AnswerLet n = 798 = a multiple of 11 + [7 + 8 - 9] 798 = a multiple of 11 + 6
$\therefore$ Remainder = 6
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