A five-digit number AABAA is divisible
Question: A five-digit number AABAA is divisible by 33. Write all the numbers of this form. Solution: Given, a number of the form AABAA is divisible by 33. Then, it is also divisible by 3 and 11, as if a number a is divisible by b, then it is also divisible by each factor of b. Since, AABAA is divisible by 3, sum its digits is also divisible by 3. i.e. 4 + 4 + 8 + A + .4 = 0,3, 6,9 or 4/4 + 8 = 0, 3, 6 9, (i) From Eq. (i), we have Further, the given number is also divisible by 11, therefore (2/4...
Read More →Find the least value that must be given
Question: Find the least value that must be given to number a, so that the number 91876a2 is divisible by 8. Solution: Given, 91876a2 is divisible by 8. Since, we know that, if a number is divisible by 8, then the number formed by last 3 digits is divisible by 8. So, 6a2 is divisible by 8. Here, a can take values from 0 to 9. For a = 0, 602 is not divisible by 8. For a = 1, 612, which is not divisible by 8. For a = 3, 632 is divisible by 8. Hence, the minimum value of a is 3 to make 91876a2 divi...
Read More →The AM between two positive numbers a and b(a>b) is twice their GM.
Question: The AM between two positive numbers a and b(ab) is twice their GM. Prove that a:b $=(2+\sqrt{3}):(2-\sqrt{3})$ Solution: To prove: Prove that a:b $=(2+\sqrt{3}):(2-\sqrt{3})$ Given: Arithmetic mean is twice of Geometric mean. Formula used: (i) Arithmetic mean between a and $b=\frac{a+b}{2}$ (ii) Geometric mean between $a$ and $b=\sqrt{a b}$ AM = 2(GM) $\frac{a+b}{2}=2(\sqrt{a b})$ $\Rightarrow a+b=4(\sqrt{a b})$ Squaring both side $\Rightarrow(a+b)^{2}=16 a b \ldots(i)$ We know that $(...
Read More →If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0,
Question: If N 5 leaves remainder 3 and N 2 leaves remainder 0, then N 10 leaves remainder 4. Solution: False Explanation: Given that, when N is divided by 5, it leaves the remainder 5. (i.e) N = 5n+3 where n= 0, 1, 2, 3, Similarly, when N is divided by 2, it leaves the remainder 0. So N is an even Number. (Using divisibility test rule of 2). But in N = 5n+3, the second term is odd. So, 5n is an odd number. When you substitute n = 1, 3, 5 in 5n+3, we will get 8, 18, 28 Now, if we divide N by 10,...
Read More →If 213x 27 is divisible by 9,
Question: If 213x 27 is divisible by 9, then the value of x is 0. Solution: False Explanation: If 213 x 27 is divisible by 9, then the sum of the digits of a number is a multiple of 9. 2+1+3+x+2+7 = 15+x Then 15x must be any multiple of 9 such as 9, 18, 27 Now assume that, 15+x = 18 X = 18 -15 X = 3...
Read More →If AB + 7C = 102,
Question: If AB + 7C = 102, where B 0, C 0, then A + B + C = 14. Solution: True Explanation: From the given number, B+C is either 2 or the two digit number that gives the unit digit as 2. It is given that B 0, C 0, If B or C = 5 or 7, then A should be 3, then it becomes A+B+C = 14 Or else, if B= C = 6, and A = 2, we will get A+B+C = 14...
Read More →If AB × 4 = 192, then A + B = 7.
Question: If AB 4 = 192, then A + B = 7. Solution: False Explanation: From the given question, the value of B should be either 3 or 8. If you take the value of B is 3, it should be equal to 19. Hence, the value of B is 8. We know that, the value of A should between 0 and 9. Hence, A x 4 = 19- 3 = 16 A = 16/4 A = 4 Therefore A= 4 and B= 8 Hence, A+B = 4+8 = 12...
Read More →If a number a is divisible by b,
Question: If a number a is divisible by b, then it must be divisible by each factor of b. Solution: True Given, a is divisible by b. Let b = p1p2, where p1 and p2 are primes. Since, a is divisible by b, a is a multiple of b i.e. a = mb = a = m.p1.p2 or a=cp2=dp1, where c = mp1, d = mp2 =a is a multiple of p1 as well as p2. Hence, a is divisible by each factor b....
Read More →Number 7N+1 will leave
Question: Number 7N+1 will leave remainder 1 when divided by 7. Solution: True Given, a number of the form 7N + 1 = x (say) Here, we observe that * is a number which is one more than a multiple of 7. i.e. when x is divided by 7, it leaves the remainder 1....
Read More →Number of the form 3N + 2 will leave
Question: Number of the form 3N + 2 will leave remainder 2 when divided by 3. Solution: True Let x = 3N + 2. Then, it can be written as. x = (a multiple of 3) + 2 i.e. x is a number which is 2 more than a multiple of 3 i.e. x is a number, which when divided by 3, leaves the remainder 2....
Read More →A three-digit number abc is divisible by 6,
Question: A three-digit number abc is divisible by 6, if c is an even number and a + b + c is a multiple of 3. Solution: True If a number is divisible by 6, then it is divisible by both 2 and 3. Since, abc is divisible by 6, it is also divisible by 2 and 3. Therefore, c is an even number and the sum of digits is divisible by 3, i.e. multiple of 3....
Read More →A four-digit mmbeFabcd is divisible
Question: A four-digit mmbeFabcd is divisible by 4, if ab is divisible by 4. Solution: False As we know that, if a number is divisible by 4, then the number formed by its digits in units and tens place is divisible by 4....
Read More →A three-digit number abc is divisible
Question: A three-digit number abc is divisible by 5, if c is an even number. Solution: False By the test of divisibility by 5, we know that if a number is divisible by 5, then its ones digit will be either 0 or 5, i.e. the numbers ending with the digits 0 or 5 are divisible by 5....
Read More →A two-digit number ab is always
Question: A two-digit number ab is always divisible by 2, if b is an even number. Solution: True By the test of divisibility by 2, we know that a number is divisible by 2, if its units digit is even....
Read More →Insert four geometric means between 6 and 192.
Question: Insert four geometric means between 6 and 192. Solution: To find: Four geometric Mean Given: The numbers 6 and 192 Formula used: (i) $r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ where $n$ is the number of geometric mean Let $G_{1}, G_{2}, G_{3}$ and $G_{4}$ be the three geometric mean Then $r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ $\Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{4+1}}$ $\Rightarrow r=\left(\frac{192}{6}\right)^{\frac{1}{4+1}}$ $\Rightarrow r=(32)^{\frac{1}{5}}$ ⇒ r...
Read More →If the digit 1 is placed after a 2-digit number
Question: If the digit 1 is placed after a 2-digit number whose tens is t and ones digit is u, the new number is ______. Solution: tu1 Explanation: Assume that, the number is initially 10t+u or tu, but after adding 1 to the unit place, tu gets shifted to the one unit higher. It means that 100t+10u+1 or tu1. State whether the statements given in questions 34 to 44 are true (T) or false (F):...
Read More →If B × B = AB, then either A = 2,
Question: If B B = AB, then either A = 2, B = 5 or A = ______, B = ______. Solution: A= 1 and B= 6 Explanation: From the given equations, B can take the values between 4 and 9 When you take B = 4, 44 = 16 Then, A=1 and B = 6...
Read More →If A × 3 = 1A, then A = ______.
Question: If A 3 = 1A, then A = ______. Solution: 5 Explanation: Using the place values of the numbers, we can write it as: 10 + A = 3A 10 = 2A A = 5...
Read More →A particle moves along the curve
Question: A particle moves along the curve $y=x^{3}$. Find the points on the curve at which the $y$-coordinate changes three times more rapidly than the $x$-coordinate. Solution: According to the question, $\frac{d y}{d t}=3 \frac{d x}{d t}$ Now, $y=x^{3}$ $\Rightarrow \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}$ $\Rightarrow 3 \frac{d x}{d t}=3 x^{2} \frac{d x}{d t}$ $\Rightarrow x^{2}=1$ $\Rightarrow x=\pm 1$ Substituting $x=\pm 1$ in $y=x^{3}$, we get $y=\pm 1$ So the points are $(1,1)$ and $(-1,...
Read More →Insert three geometric means between
Question: Insert three geometric means between $\frac{1}{3}$ and 432 . Solution: To find: Three geometric Mean Given: The numbers $\frac{1}{3}$ and 432 Formula used: (i) r $=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ where n is the number of geometric mean Let $G_{1}, G_{2}$ and $G_{3}$ be the three geometric mean Then $r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$ $\Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{3+1}}$ $\Rightarrow r=\left(\frac{432}{\left(\frac{1}{3}\right)}\right)^{\frac{1}{2+1...
Read More →If o three-digit number abc
Question: If o three-digit number abc is divisible by 11, thenis either 0 or multiple of 11. Solution: (a+c)-b Since, abc is divisible by 11, the difference of sum of its digits at odd places and that of even places is either zero or multiple of 11, i.e. (a + c) b is either zero or multiple of 11....
Read More →A number is divisible by 11,
Question: A number is divisible by 11, if the differences between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by -. Solution: 11 By test of divisibility by 1,1, we know that, a number is divisible by 11, if the sum of digits at odd places and even places are equal or differ by a number, which is divisible by 11....
Read More →If the solve the problem
Question: If $y=7 x-x^{3}$ and $x$ increases at the rate of 4 units per second, how fast is the slope of the curve changing when $x=2$ ? Solution: Here, $y=7 x-x^{3}$ $\Rightarrow \frac{d y}{d x}=7 x-x^{3}$ Let $s$ be the slope. Then, $s=7-3 x^{2}$ $\Rightarrow \frac{d s}{d t}=-6 x \frac{d x}{d t}$ $\Rightarrow \frac{d s}{d t}=-6(4)(2)$ $\left[\because x=2\right.$ and $\frac{d x}{d t}=4$ units / sec $]$ $\Rightarrow \frac{d s}{d t}=-48$...
Read More →A four-digit number abed
Question: A four-digit number abed is divisible by 11, if d + b =or-. Solution: a + c,12(a + c) We know that, a number is divisible by 11, if the difference between the sum of digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11. Hence, abcd is divisible by 11, if (d + b)- (a + c) = 0,11,22, 33, = d + b = a + c or d + b = 12(a + c)...
Read More →1 x 35 is divisible by 9 if x = _______.
Question: 1 x 35 is divisible by 9 if x = _______. Solution: 0 Explanation: We know that, when a number is a multiple of 9, then the sum of digits in the number is divisible by 9 So, by adding the digits, we get: X+9 = 9 Hence, x= 0...
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